JEE Vector Algebra Questions

Let $$\vec{a}$$ and $$\vec{b}$$ be unit vectors with the angle $$\frac{\pi}{3}$$ between them, and consider the vectors $$\lambda\vec{a} + 2\vec{b}$$ and $$3\vec{a} - \lambda\vec{b}$$ for $$\lambda\in[-1,3]$$; we seek the number of values of $$\lambda$$ for which these vectors are perpendicular.

Since perpendicularity means their dot product vanishes, we write

$$(\lambda\vec{a} + 2\vec{b}) \cdot (3\vec{a} - \lambda\vec{b}) = 0\,. $$

Expanding the dot product gives

$$3\lambda(\vec{a}\cdot\vec{a}) - \lambda^2(\vec{a}\cdot\vec{b}) + 6(\vec{b}\cdot\vec{a}) - 2\lambda(\vec{b}\cdot\vec{b}) = 0\,. $$

Because $$|\vec{a}|=|\vec{b}|=1$$ we have $$\vec{a}\cdot\vec{a}=1$$ and $$\vec{b}\cdot\vec{b}=1$$, while $$\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\frac{\pi}{3}=\frac12\,. $$

Substituting these results into the expanded dot product yields

$$3\lambda - \frac{\lambda^2}{2} + 6\cdot\frac12 - 2\lambda = 0\,, $$

which simplifies to

$$\lambda - \frac{\lambda^2}{2} + 3 = 0\,. $$

Multiplying both sides by $$-2$$ gives the quadratic equation

$$\lambda^2 - 2\lambda - 6 = 0\,. $$

Solving this quadratic leads to

$$\lambda = \frac{2\pm\sqrt{4+24}}{2} = 1\pm\sqrt{7}\,. $$

Since $$\sqrt{7}\approx2.646$$, the two solutions are approximately $$1+\sqrt{7}\approx3.646$$ and $$1-\sqrt{7}\approx-1.646$$, neither of which lies in the interval $$[-1,3]$$.

Therefore there are no values of $$\lambda$$ in $$[-1,3]$$ that make the given vectors perpendicular, and hence the number of such values is 0.

Final Answer: 0.

Arc AC subtends $$90°$$ at center O. B divides arc AC in ratio 1:5, so arc AB subtends $$\frac{1}{6} \times 90° = 15°$$.

With A at angle $$0°$$, B at $$15°$$, C at $$90°$$ (unit radius):

$$\vec{OA} = (1, 0)$$, $$\vec{OB} = (\cos 15°, \sin 15°)$$, $$\vec{OC} = (0, 1)$$.

From $$\vec{OC} = \alpha\vec{OA} + \beta\vec{OB}$$:

$$0 = \alpha + \beta\cos 15°$$ and $$1 = \beta\sin 15°$$.

$$\beta = \frac{1}{\sin 15°} = \frac{4}{\sqrt{6}-\sqrt{2}} = \sqrt{6}+\sqrt{2}$$

$$\alpha = -\cot 15° = -(2+\sqrt{3})$$

$$(\sqrt{3}-1)(\sqrt{6}+\sqrt{2}) = 3\sqrt{2}-\sqrt{2} = 2\sqrt{2}$$

$$\sqrt{2}(\sqrt{3}-1)\beta = \sqrt{2} \times 2\sqrt{2} = 4$$

$$\alpha + \sqrt{2}(\sqrt{3}-1)\beta = -(2+\sqrt{3}) + 4 = 2 - \sqrt{3}$$

Let $$\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$$ be a nonzero vector. The projection of $$\vec{a}$$ on a vector $$\vec{v}$$ is $$\frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}$$.

The three given vectors are $$\vec{v_1} = 2\hat{i} - \hat{j} + 2\hat{k}$$, $$\vec{v_2} = \hat{i} + 2\hat{j} - 2\hat{k}$$, and $$\vec{v_3} = \hat{k}$$.

Their magnitudes are: $$|\vec{v_1}| = \sqrt{4+1+4} = 3$$, $$|\vec{v_2}| = \sqrt{1+4+4} = 3$$, $$|\vec{v_3}| = 1$$.

The projections are:

$$\text{proj}_1 = \frac{2x - y + 2z}{3}, \quad \text{proj}_2 = \frac{x + 2y - 2z}{3}, \quad \text{proj}_3 = z$$

Setting proj$$_1$$ = proj$$_2$$: $$2x - y + 2z = x + 2y - 2z$$, which gives $$x - 3y + 4z = 0$$ ...(1)

Setting proj$$_2$$ = proj$$_3$$: $$\frac{x + 2y - 2z}{3} = z$$, which gives $$x + 2y - 5z = 0$$ ...(2)

From (1) and (2): subtracting (2) from (1): $$-5y + 9z = 0$$, so $$y = \frac{9z}{5}$$.

Substituting into (2): $$x + \frac{18z}{5} - 5z = 0$$, so $$x = 5z - \frac{18z}{5} = \frac{7z}{5}$$.

Taking $$z = 5$$: $$\vec{a} = 7\hat{i} + 9\hat{j} + 5\hat{k}$$.

$$|\vec{a}| = \sqrt{49 + 81 + 25} = \sqrt{155}$$.

The unit vector along $$\vec{a}$$ is $$\frac{1}{\sqrt{155}}(7\hat{i} + 9\hat{j} + 5\hat{k})$$.

Hence, the correct answer is Option C.

Let the vertices of the triangle be $$P = 4\vec{p} + \vec{q} - 3\vec{r}$$, $$Q = -5\vec{p} + \vec{q} + 2\vec{r}$$, $$R = 2\vec{p} - \vec{q} + 2\vec{r}$$.

The orthocenter is $$H = \frac{\vec{p} + \vec{q} + \vec{r}}{4}$$.

We use the property that the centroid $$G$$ divides $$OH$$ in ratio $$1:2$$ (where $$O$$ is circumcenter), i.e., $$G$$ divides segment from circumcenter to orthocenter. More precisely: $$\vec{G} = \frac{\vec{P}+\vec{Q}+\vec{R}}{3}$$.

$$\vec{G} = \frac{(4-5+2)\vec{p} + (1+1-1)\vec{q} + (-3+2+2)\vec{r}}{3} = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$$

Euler line relation: $$\vec{G} = \frac{\vec{H} + 2\vec{O}}{3}$$ where $$O$$ is the circumcenter.

$$\frac{\vec{p}+\vec{q}+\vec{r}}{3} = \frac{1}{3}\left(\frac{\vec{p}+\vec{q}+\vec{r}}{4} + 2\vec{O}\right)$$

$$\vec{p}+\vec{q}+\vec{r} = \frac{\vec{p}+\vec{q}+\vec{r}}{4} + 2\vec{O}$$

$$2\vec{O} = \frac{3(\vec{p}+\vec{q}+\vec{r})}{4}$$

$$\vec{O} = \frac{3}{8}(\vec{p}+\vec{q}+\vec{r}) = \frac{3}{8}\vec{p} + \frac{3}{8}\vec{q} + \frac{3}{8}\vec{r}$$

So $$\alpha = 3/8$$, $$\beta = 3/8$$, $$\gamma = 3/8$$.

$$\alpha + 2\beta + 5\gamma = \frac{3}{8} + \frac{6}{8} + \frac{15}{8} = \frac{24}{8} = 3$$.

The correct answer is Option A: 3.

The vectors lie in a plane if and only if their scalar triple product is zero, i.e. $$\vec{X}\cdot(\vec{Y}\times\vec{Z}) = 0$$. In component form this is equivalent to the determinant of the three column vectors being zero.

First write each required vector in $$\hat i,\hat j,\hat k$$ components.

$$\vec{x} = (1,\,2,\,3),\; \vec{y} = (2,\,3,\,1),\; \vec{z} = (3,\,1,\,2)$$

$$\vec{X}= \alpha\vec{x}+\beta\vec{y}-\vec{z} = (\alpha+2\beta-3,\; 2\alpha+3\beta-1,\; 3\alpha+\beta-2)$$

$$\vec{Y}= \alpha\vec{y}+\beta\vec{z}-\vec{x} = (2\alpha+3\beta-1,\; 3\alpha+\beta-2,\; \alpha+2\beta-3)$$

$$\vec{Z}= \alpha\vec{z}+\beta\vec{x}-\vec{y} = (3\alpha+\beta-2,\; \alpha+2\beta-3,\; 2\alpha+3\beta-1)$$

Introduce the shorthand

$$A = \alpha+2\beta-3,\qquad B = 2\alpha+3\beta-1,\qquad C = 3\alpha+\beta-2$$

Then$$ \vec{X} = (A,\;B,\;C),\; \vec{Y} = (B,\;C,\;A),\; \vec{Z} = (C,\;A,\;B). $$

Set up the determinant:

$$ \left| \begin{array}{ccc} A & B & C\\ B & C & A\\ C & A & B \end{array} \right| = 0. $$

Evaluating the determinant,

$$ \begin{aligned} D &= A(CB - A^{2}) - B(B^{2} - AC) + C(AB - C^{2}) \\ &= (ABC - A^{3}) + (ABC - B^{3}) + (ABC - C^{3}) \\ &= 3ABC - (A^{3}+B^{3}+C^{3}). \end{aligned} $$

Hence$$D = 0 \Longrightarrow A^{3}+B^{3}+C^{3}-3ABC = 0.$$ Using the identity$$ A^{3}+B^{3}+C^{3}-3ABC = (A+B+C)\,\bigl(A^{2}+B^{2}+C^{2}-AB-BC-CA\bigr), $$we get two possibilities:

$$A+B+C = 0.$$

$$A=B=C.$$

Because $$\alpha,\beta$$ are distinct positive numbers, Case 2 is impossible (solving $$A=B$$ forces $$\alpha+\beta=-2$$, which contradicts positivity). Therefore only Case 1 survives.

Compute $$A+B+C$$:

$$ \begin{aligned} A+B+C &= (\alpha+2\beta-3) + (2\alpha+3\beta-1) + (3\alpha+\beta-2)\\ &= 6\alpha + 6\beta - 6\\ &= 6(\alpha+\beta-1). \end{aligned} $$

Setting this to zero gives $$\alpha+\beta-1 = 0 \Longrightarrow \alpha+\beta = 1.$$ Both $$\alpha,\beta$$ can be positive and distinct while satisfying this relation, so the condition is admissible.

Finally, the required expression is

$$\alpha+\beta-3 = 1 - 3 = -2.$$

Hence the answer is -2.

Given $$\mathbf{a} = 3\hat{i} - \hat{j} + 2\hat{k}$$ and $$\mathbf{b} = \mathbf{a} \times (\hat{i} - 2\hat{k})$$:

$$\mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{vmatrix}$$ $$\mathbf{b} = \hat{i}(2 - 0) - \hat{j}(-6 - 2) + \hat{k}(0 + 1)$$ $$\mathbf{b} = 2\hat{i} + 8\hat{j} + \hat{k}$$

Given $$\mathbf{c} = \mathbf{b} \times \hat{k}$$:

$$\mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 8 & 1 \\ 0 & 0 & 1 \end{vmatrix}$$ $$\mathbf{c} = \hat{i}(8 - 0) - \hat{j}(2 - 0) + \hat{k}(0 - 0)$$ $$\mathbf{c} = 8\hat{i} - 2\hat{j}$$

$$\mathbf{c} - 2\hat{j} = (8\hat{i} - 2\hat{j}) - 2\hat{j} = 8\hat{i} - 4\hat{j}$$

The formula for the projection of a vector $$\mathbf{u}$$ onto a vector $$\mathbf{v}$$ is $$\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|}$$.

First, let's find the magnitude of $$\mathbf{a}$$:

$$|\mathbf{a}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$$

$$(\mathbf{c} - 2\hat{j}) \cdot \mathbf{a}$$ = $$(8\hat{i} - 4\hat{j} + 0\hat{k}) \cdot (3\hat{i} - \hat{j} + 2\hat{k})$$= (8)(3) + (-4)(-1) + (0)(2)= 28

$$\text{Projection} = \frac{28}{\sqrt{14}} = \frac{2 \times 14}{\sqrt{14}} = 2\sqrt{14}$$

The given vectors are $$\vec u = 3\hat i - \hat j$$ and $$\vec v = 2\hat i + \hat j - \lambda \hat k$$ with $$\lambda \gt 0$$.
The angle $$\theta$$ between them satisfies

$$\cos\theta = \dfrac{\vec u \cdot \vec v}{\lvert\vec u\rvert\,\lvert\vec v\rvert} = \dfrac{\sqrt5}{2\sqrt7}\,\,\,\,-(1)$$

First compute the numerator:
$$\vec u \cdot \vec v = (3)(2) + (-1)(1) + (0)(-\lambda) = 6 - 1 = 5$$

Next compute the magnitudes:
$$\lvert\vec u\rvert = \sqrt{3^{2} + (-1)^{2}} = \sqrt{9 + 1} = \sqrt{10}$$ $$\lvert\vec v\rvert = \sqrt{2^{2} + 1^{2} + \lambda^{2}} = \sqrt{5 + \lambda^{2}}$$

Insert these results in $$(1)$$:

$$\frac{5}{\sqrt{10}\,\sqrt{5+\lambda^{2}}} = \frac{\sqrt5}{2\sqrt7}$$

Cross-multiply and simplify:
$$10\sqrt7 = \sqrt5\,\sqrt{10}\,\sqrt{5+\lambda^{2}}$$ $$10\sqrt7 = (\sqrt{5\cdot10})\,\sqrt{5+\lambda^{2}} = 5\sqrt2\,\sqrt{5+\lambda^{2}}$$ Divide by $$5$$:
$$2\sqrt7 = \sqrt2\,\sqrt{5+\lambda^{2}}$$ Square both sides:
$$(2\sqrt7)^{2} = (\sqrt2\,\sqrt{5+\lambda^{2}})^{2}$$ $$4\cdot7 = 2\,(5+\lambda^{2})$$ $$28 = 10 + 2\lambda^{2}$$ $$18 = 2\lambda^{2}$$ $$\lambda^{2} = 9 \Longrightarrow \lambda = 3 \quad(\text{since } \lambda \gt 0)$$

Thus $$\vec v = 2\hat i + \hat j - 3\hat k$$ and its magnitude is

$$\lvert\vec v\rvert^{2} = 2^{2} + 1^{2} + (-3)^{2} = 4 + 1 + 9 = 14\,\,\,\,-(2)$$

Now write $$\vec v = \vec v_1 + \vec v_2$$ where $$\vec v_1$$ is parallel to $$\vec u$$ and $$\vec v_2$$ is perpendicular to $$\vec u$$.
Because a parallel component and a perpendicular component are orthogonal, we have the Pythagorean relation

$$\lvert\vec v\rvert^{2} = \lvert\vec v_1\rvert^{2} + \lvert\vec v_2\rvert^{2}\,\,\,\,-(3)$$

Substituting the value from $$(2)$$ into $$(3)$$ gives

$$\lvert\vec v_1\rvert^{2} + \lvert\vec v_2\rvert^{2} = 14$$

Hence the required value is $$14$$.
Option B is correct.

The three vectors are connected by
$$\vec{u}\times\vec{v}= \vec{w},\qquad \vec{v}\times\vec{w}= \vec{u},\qquad \vec{w}= \hat{i}+ \hat{j}-2\hat{k}.$$ The cross-product of two vectors is perpendicular to each of them, so

$$\vec{u}\cdot\vec{v}=0, \qquad \vec{v}\cdot\vec{w}=0, \qquad \vec{w}\cdot\vec{u}=0.$$

Hence the three vectors are pair-wise perpendicular.

Magnitude of $$\vec{w}$$:
$$|\vec{w}|^{2}=1^{2}+1^{2}+(-2)^{2}=6.$$

Using the identity $$|\vec{a}\times\vec{b}|^{2}=|\,\vec{a}\,|^{2}|\,\vec{b}\,|^{2}-(\vec{a}\cdot\vec{b})^{2},$$
$$|\vec{w}|^{2}=|\vec{u}\times\vec{v}|^{2}=|\vec{u}|^{2}|\vec{v}|^{2}\quad(\because\vec{u}\cdot\vec{v}=0)$$ and
$$|\vec{u}|^{2}=|\vec{v}\times\vec{w}|^{2}=|\vec{v}|^{2}|\vec{w}|^{2}\quad(\because\vec{v}\cdot\vec{w}=0).$$

Let $$|\vec{u}|^{2}=a,\; |\vec{v}|^{2}=b.$$
Then $$6=ab \quad\text{and}\quad a=6b.$$ Substituting, $$6=(6b)b\;\Longrightarrow\; b^{2}=1\;\Longrightarrow\; b=1.$$ Therefore

Case P: $$|\vec{v}|^{2}=1\; \Rightarrow\; (P)\to(2).$$

The components of $$\vec{u}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$$ obey the three equations

$$-t\alpha+\beta+\gamma=0\; -(1)$$
$$\alpha-t\beta+\gamma=0\; -(2)$$
$$\alpha+\beta-t\gamma=0\; -(3).$$

For a non-trivial solution, the determinant of the coefficient matrix must vanish:

$$\begin{vmatrix}-t&1&1\\[2pt]1&-t&1\\[2pt]1&1&-t\end{vmatrix}=0 \;\Longrightarrow\; -t^{3}+3t+2=0 \;\Longrightarrow\; t^{3}-3t-2=0 =(t-2)(t+1)^{2}=0.$$

Thus $$t=2\quad\text{or}\quad t=-1.$$ We must also satisfy $$\vec{w}\cdot\vec{u}=0\;\Rightarrow\; \alpha+\beta-2\gamma=0\; -(4).$$

Taking $$t=-1$$
Equations (1)-(3) all reduce to $$\alpha+\beta+\gamma=0.$$ With $$\alpha=\sqrt{3}$$, $$\beta+\gamma=-\sqrt{3}\; -(5).$$ Using (4): $$\sqrt{3}+\beta-2\gamma=0 \;\Longrightarrow\; -3\gamma=0 \;\Longrightarrow\; \gamma=0,\;\beta=-\sqrt{3}.$$ Hence

Case Q: $$\gamma^{2}=0\;\Rightarrow\; (Q)\to(1).$$

Case R: $$(\beta+\gamma)^{2}=(-\sqrt{3})^{2}=3\;\Rightarrow\; (R)\to(4).$$

Taking $$t=2$$
From (1): $$\beta=2\alpha-\gamma.$$ From (2): $$\alpha-2\beta+\gamma=0 \;\Longrightarrow\; \gamma=\alpha.$$ Therefore $$\beta=\alpha.$$ With $$\alpha=\sqrt{2}$$ we get $$t=2,$$ hence

Case S: $$t+3=2+3=5\;\Rightarrow\; (S)\to(5).$$

Collecting all matches:
(P)$$\to$$(2), (Q)$$\to$$(1), (R)$$\to$$(4), (S)$$\to$$(5).

Thus the correct option is
Option A which is: (P)$$\to$$(2), (Q)$$\to$$(1), (R)$$\to$$(4), (S)$$\to$$(5).

Let $$\vec{a}=2\hat{i}-3\hat{j}+\hat{k}$$, $$\vec{b}=3\hat{i}+2\hat{j}+5\hat{k}$$ and $$\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$$.

The condition $$(\vec{a}-\vec{c})\times\vec{b}=-18\hat{i}-3\hat{j}+12\hat{k}$$ gives three scalar equations.
Write $$\vec{a}-\vec{c}=(2-x)\hat{i}+(-3-y)\hat{j}+(1-z)\hat{k}.$$

For two vectors $$\vec{p}=(p_1,p_2,p_3)$$ and $$\vec{q}=(q_1,q_2,q_3)$$, the cross product is
$$\vec{p}\times\vec{q}=\bigl(p_2q_3-p_3q_2\bigr)\hat{i}-\bigl(p_1q_3-p_3q_1\bigr)\hat{j}+\bigl(p_1q_2-p_2q_1\bigr)\hat{k}.$$

Putting $$\vec{p}=(2-x,-3-y,1-z)$$ and $$\vec{q}=(3,2,5)$$:

$$\begin{aligned} (\vec{a}-\vec{c})\times\vec{b} &=\bigl[5(-3-y)-2(1-z)\bigr]\hat{i}\\ &\;\;-\bigl[5(2-x)-3(1-z)\bigr]\hat{j}\\ &\;\;+\bigl[2(2-x)-3(-3-y)\bigr]\hat{k}. \end{aligned}$$

Equating with $$-18\hat{i}-3\hat{j}+12\hat{k}$$ gives

$$5(-3-y)-2(1-z)=-18$$ $$-(1)$$
$$5(2-x)-3(1-z)=3$$ $$-(2)$$
$$2(2-x)-3(-3-y)=12$$ $$-(3)$$

Simplify each:

From $$(1):\; -15-5y-2+2z=-18 \;\Rightarrow\; 5y-2z=1 \;-(4)$$

From $$(2):\; 10-5x-3+3z=3 \;\Rightarrow\; 5x-3z=4 \;-(5)$$

From $$(3):\; 4-2x+9+3y=12 \;\Rightarrow\; 2x-3y=1 \;-(6)$$

The second given condition is $$\vec{a}\cdot\vec{c}=3$$:
$$2x-3y+z=3 \;-(7)$$

Solve equations $$(4)-(7).$$

From $$(6):\; 2x-3y=1 \;\Rightarrow\; y=\frac{2x-1}{3}.$$ Substitute this in $$(7):$$
$$2x-3\!\left(\frac{2x-1}{3}\right)+z=3$$
$$\Rightarrow\;2x-(2x-1)+z=3 \;\Rightarrow\;1+z=3 \;\Rightarrow\; z=2.$$

Insert $$z=2$$ in $$(5):$$
$$5x-3(2)=4 \;\Rightarrow\; 5x-6=4 \;\Rightarrow\; x=2.$$

Put $$x=2$$ in $$(6):$$
$$2(2)-3y=1 \;\Rightarrow\; 4-3y=1 \;\Rightarrow\; y=1.$$

Hence $$\vec{c}=2\hat{i}+\hat{j}+2\hat{k}.$$

Now find $$\vec{d}=\vec{b}\times\vec{c}.$$ Compute the determinant:
$$\vec{d}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 2 & 5\\ 2 & 1 & 2 \end{vmatrix} = \bigl(2\cdot2-5\cdot1\bigr)\hat{i}-\bigl(3\cdot2-5\cdot2\bigr)\hat{j}+\bigl(3\cdot1-2\cdot2\bigr)\hat{k}.$$

Simplifying:
$$\vec{d}=-1\hat{i}+4\hat{j}-1\hat{k}= -\hat{i}+4\hat{j}-\hat{k}.$$

Finally compute $$\vec{a}\cdot\vec{d}:$$
$$\vec{a}\cdot\vec{d}=(2,-3,1)\cdot(-1,4,-1)=-2-12-1=-15.$$ Therefore
$$\lvert\vec{a}\cdot\vec{d}\rvert=15.$$

Option D is correct.

Let the common magnitude of the two vectors be $$|\vec a| = |\vec b| = m$$ and let the angle between them be $$\theta$$.

Using the cosine rule for vectors:
$$|\vec a + \vec b| = \sqrt{m^2 + m^2 + 2m^2 \cos\theta} = m\sqrt{2(1+\cos\theta)}$$
$$|\vec a - \vec b| = \sqrt{m^2 + m^2 - 2m^2 \cos\theta} = m\sqrt{2(1-\cos\theta)}$$

Rewrite these with the half-angle substitution $$\theta = 2x$$ (so $$x = \theta/2$$):
$$|\vec a + \vec b| = 2m\cos x$$
$$|\vec a - \vec b| = 2m\sin x$$

The given ratio becomes
$$\frac{|\vec a+\vec b| + |\vec a-\vec b|}{|\vec a+\vec b| - |\vec a-\vec b|} = \frac{2m(\cos x + \sin x)}{2m(\cos x - \sin x)} = \frac{\cos x + \sin x}{\cos x - \sin x}$$

This ratio equals $$\sqrt{2}+1$$, hence
$$\frac{\cos x + \sin x}{\cos x - \sin x} = \sqrt{2}+1 \; -(1)$$

Cross-multiplying in $$(1)$$:
$$\cos x + \sin x = (\sqrt{2}+1)(\cos x - \sin x)$$
$$\cos x + \sin x - (\sqrt{2}+1)\cos x + (\sqrt{2}+1)\sin x = 0$$
$$\cos x(1-\sqrt{2}-1) + \sin x(1+\sqrt{2}+1) = 0$$
$$\cos x(-\sqrt{2}) + \sin x(\sqrt{2}+2) = 0$$
$$\tan x = \frac{\sqrt{2}}{\sqrt{2}+2} \; -(2)$$

The quantity required is
$$\frac{|\vec a+\vec b|^{2}}{|\vec a|^{2}} = \frac{(2m\cos x)^{2}}{m^{2}} = 4\cos^{2}x \; -(3)$$

From $$(2)$$, set $$\tan x = t = \dfrac{\sqrt{2}}{\sqrt{2}+2}$$. Then
$$\cos^{2}x = \frac{1}{1+t^{2}}$$

Compute $$t^{2}$$:
$$t^{2} = \frac{2}{(\sqrt{2}+2)^{2}} = \frac{2}{6+4\sqrt{2}} = \frac{1}{3+2\sqrt{2}}$$

Hence
$$1+t^{2} = 1 + \frac{1}{3+2\sqrt{2}} = \frac{4+2\sqrt{2}}{3+2\sqrt{2}}$$

Insert this into $$(3)$$:
$$4\cos^{2}x = \frac{4}{1+t^{2}} = 4 \cdot \frac{3+2\sqrt{2}}{4+2\sqrt{2}}$$
$$= \frac{12 + 8\sqrt{2}}{4 + 2\sqrt{2}} = \frac{6 + 4\sqrt{2}}{2 + \sqrt{2}} = 2 + \sqrt{2}$$

Therefore,
$$\frac{|\vec a + \vec b|^{2}}{|\vec a|^{2}} = 2 + \sqrt{2}$$

The correct choice is Option C.

First, find a vector perpendicular to both $$\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$$ and $$\vec{c} = 2\hat{i} + 3\hat{j} - \hat{k}$$.

$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix}$$ which simplifies to $$= \hat{i}(2-9) - \hat{j}(-1-6) + \hat{k}(3+4) = -7\hat{i} + 7\hat{j} + 7\hat{k}$$.

Therefore, $$\vec{b} \times \vec{c} = 7(-\hat{i} + \hat{j} + \hat{k})$$.
The unit vector in this direction is given by $$\hat{a} = \pm\frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k})$$.

Next, $$\hat{a}$$ makes an angle $$\cos^{-1}\bigl(-\tfrac{1}{3}\bigr)$$ with $$\hat{i} + \hat{j} + \hat{k}$$, so

$$\hat{a} \cdot (\hat{i} + \hat{j} + \hat{k}) = |\hat{a}|\;|\hat{i}+\hat{j}+\hat{k}|\;\cos\theta.$$
For $$\hat{a} = \frac{1}{\sqrt{3}}(-1+1+1) = \frac{1}{\sqrt{3}},$$

$$\cos\theta = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}.$$
For $$\hat{a} = \frac{-1}{\sqrt{3}}(-1+1+1) = \frac{-1}{\sqrt{3}},$$
$$\cos\theta = \frac{-1/\sqrt{3}}{\sqrt{3}} = -\frac{1}{3}$$ 

Hence, $$\hat{a} = \frac{-1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k}) = \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} - \hat{k}).$$

Now, $$\hat{a}$$ makes an angle $$\frac{\pi}{3}$$ with $$\hat{i} + \alpha\hat{j} + \hat{k}$$, so
$$\frac{1}{2} = \frac{-\alpha}{\sqrt{3(2+\alpha^2)}}.$$

Since the left side is positive, $$\alpha$$ must be negative. Squaring both sides yields
$$\frac{1}{4} = \frac{\alpha^2}{3(2+\alpha^2)},$$
$$3(2+\alpha^2) = 4\alpha^2,\quad 6 + 3\alpha^2 = 4\alpha^2,\quad \alpha^2 = 6,\quad \alpha = -\sqrt{6}.$$

$$\vec{a} = \hat{i}+2\hat{j}+3\hat{k}$$, $$\vec{b} = 3\hat{i}+\hat{j}-\hat{k}$$. $$\vec{c}$$ is coplanar with $$\vec{a}$$ and $$\vec{b}$$, perpendicular to $$\vec{b}$$, and $$\vec{a}\cdot\vec{c} = 5$$.

Since $$\vec{c}$$ is coplanar with $$\vec{a}$$ and $$\vec{b}$$: $$\vec{c} = \lambda\vec{a} + \mu\vec{b}$$

$$\vec{c} \perp \vec{b}$$: $$\vec{c}\cdot\vec{b} = 0$$

$$\lambda(\vec{a}\cdot\vec{b}) + \mu(\vec{b}\cdot\vec{b}) = 0$$

$$\vec{a}\cdot\vec{b} = 3+2-3 = 2$$, $$|\vec{b}|^2 = 9+1+1 = 11$$

$$2\lambda + 11\mu = 0 \Rightarrow \lambda = -\frac{11\mu}{2}$$

$$\vec{a}\cdot\vec{c} = 5$$

$$\lambda|\vec{a}|^2 + \mu(\vec{a}\cdot\vec{b}) = 5$$

$$|\vec{a}|^2 = 1+4+9 = 14$$

$$14\lambda + 2\mu = 5$$

$$14(-11\mu/2) + 2\mu = 5$$

$$-77\mu + 2\mu = 5$$

$$-75\mu = 5 \Rightarrow \mu = -1/15$$

$$\lambda = -11(-1/15)/2 = 11/30$$

$$\vec{c} = \frac{11}{30}(1,2,3) - \frac{1}{15}(3,1,-1)$$

$$= (\frac{11}{30}-\frac{2}{10}, \frac{22}{30}-\frac{1}{15}, \frac{33}{30}+\frac{1}{15})$$

$$= (\frac{11-6}{30}, \frac{22-2}{30}, \frac{33+2}{30}) = (\frac{5}{30}, \frac{20}{30}, \frac{35}{30}) = (\frac{1}{6}, \frac{2}{3}, \frac{7}{6})$$

$$|\vec{c}|^2 = \frac{1}{36} + \frac{4}{9} + \frac{49}{36} = \frac{1+16+49}{36} = \frac{66}{36} = \frac{11}{6}$$

$$|\vec{c}| = \sqrt{\frac{11}{6}}$$

The correct answer is Option 1: $$\sqrt{\frac{11}{6}}$$.

$$\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$$, $$\vec{b} = \lambda\hat{i} + 4\hat{k}$$ ($$\lambda > 0$$). $$\vec{c}$$ is the projection vector of $$\vec{b}$$ on $$\vec{a}$$. $$|\vec{a} + \vec{c}| = 7$$.

The projection of $$\vec{b}$$ onto $$\vec{a}$$ is:

$$\vec{c} = \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a}$$

$$\vec{b} \cdot \vec{a} = \lambda(1) + 0(2) + 4(2) = \lambda + 8$$

$$|\vec{a}|^2 = 1 + 4 + 4 = 9$$

$$\vec{c} = \frac{\lambda + 8}{9} \vec{a} = \frac{\lambda + 8}{9}(\hat{i} + 2\hat{j} + 2\hat{k})$$

$$\vec{a} + \vec{c} = \left(1 + \frac{\lambda+8}{9}\right)\vec{a} = \frac{9 + \lambda + 8}{9}\vec{a} = \frac{\lambda + 17}{9}\vec{a}$$

$$|\vec{a} + \vec{c}| = \frac{|\lambda + 17|}{9} |\vec{a}| = \frac{\lambda + 17}{9} \times 3 = \frac{\lambda + 17}{3}$$

(Since $$\lambda > 0$$, $$\lambda + 17 > 0$$)

$$\frac{\lambda + 17}{3} = 7 \implies \lambda = 4$$

$$\vec{b} = 4\hat{i} + 4\hat{k}$$

$$\vec{c} = \frac{4+8}{9}(\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{12}{9}(\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{4}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$$

$$\vec{c} = \frac{4}{3}\hat{i} + \frac{8}{3}\hat{j} + \frac{8}{3}\hat{k}$$

Area = $$|\vec{b} \times \vec{c}|$$

$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 0 & 4 \\ 4/3 & 8/3 & 8/3 \end{vmatrix}$$

$$= \hat{i}(0 \cdot \frac{8}{3} - 4 \cdot \frac{8}{3}) - \hat{j}(4 \cdot \frac{8}{3} - 4 \cdot \frac{4}{3}) + \hat{k}(4 \cdot \frac{8}{3} - 0)$$

$$= \hat{i}(-\frac{32}{3}) - \hat{j}(\frac{16}{3}) + \hat{k}(\frac{32}{3})$$

$$|\vec{b} \times \vec{c}| = \frac{1}{3}\sqrt{32^2 + 16^2 + 32^2} = \frac{1}{3}\sqrt{1024 + 256 + 1024} = \frac{1}{3}\sqrt{2304} = \frac{48}{3} = 16$$

The area is 16.

Given
$$\vec{a}= \hat{i}+ \hat{j}+ \hat{k},\; \vec{b}=3\hat{i}+2\hat{j}-\hat{k},\; \vec{c}= \lambda\hat{j}+ \mu\hat{k},\; \hat{d}\; \text{(unit)}$$
with the conditions
$$\vec{a}\times\hat{d}=\vec{b}\times\hat{d},\qquad \vec{c}\cdot\hat{d}=1,\qquad \vec{c}\perp\vec{a}.$$

Step 1: Relating $$\vec{a}$$ and $$\vec{b}$$ to $$\hat{d}$$
The identity $$\vec{u}\times\hat{d}=\vec{v}\times\hat{d}\;\Longrightarrow\;(\vec{u}-\vec{v})\times\hat{d}= \vec{0}$$ means $$\vec{u}-\vec{v}$$ is parallel to $$\hat{d}$$.
Thus $$\bigl(\vec{a}-\vec{b}\bigr)\times\hat{d}= \vec{0}\;\Longrightarrow\;\vec{a}-\vec{b}=t\hat{d}\;$$ for some scalar $$t$$.

Compute $$\vec{a}-\vec{b}$$:
$$\vec{a}-\vec{b}=(1-3)\hat{i}+(1-2)\hat{j}+(1-(-1))\hat{k}=-2\hat{i}-\hat{j}+2\hat{k}.$$

The magnitude of this vector is
$$\lvert\vec{a}-\vec{b}\rvert=\sqrt{(-2)^2+(-1)^2+2^2}=3.$$

Therefore the required unit vector is
$$\hat{d}=s\,\frac{-2\hat{i}-\hat{j}+2\hat{k}}{3},\qquad s=\pm1.$$
Explicitly
$$\hat{d}=s\Bigl(-\tfrac23\,\hat{i}-\tfrac13\,\hat{j}+\tfrac23\,\hat{k}\Bigr).$$

Step 2: Using $$\vec{c}\cdot\hat{d}=1$$
With $$\vec{c}=(0,\lambda,\mu)$$, dot product gives
$$\vec{c}\cdot\hat{d}=s\Bigl(-\tfrac{\lambda}{3}+\tfrac{2\mu}{3}\Bigr)=1$$
$$\Longrightarrow\;s(-\lambda+2\mu)=3\;.$$ $$-(1)$$

Step 3: Using $$\vec{c}\perp\vec{a}$$
Orthogonality with $$\vec{a}=(1,1,1)$$ yields
$$\vec{a}\cdot\vec{c}=0\;\Longrightarrow\;\lambda+\mu=0\;\Longrightarrow\;\mu=-\lambda.$$ $$-(2)$$

Step 4: Determining $$\lambda,\mu$$
Substituting $$\mu=-\lambda$$ from $$(2)$$ into $$(1)$$:
$$s(-\lambda+2(-\lambda))=s(-3\lambda)=3\;\Longrightarrow\;\lambda=-\frac{1}{s},\qquad \mu=-\lambda=\frac{1}{s}.$$

Hence
For $$s=1:\; \lambda=-1,\;\mu=1,\;\hat{d}=(-\tfrac23,-\tfrac13,\tfrac23).$$
For $$s=-1:\;\lambda=1,\;\mu=-1,\;\hat{d}=(\tfrac23,\tfrac13,-\tfrac23).$$

Step 5: Evaluating $$\bigl|\,3\lambda\hat{d}+\mu\vec{c}\bigr|^2$$
Take the case $$s=1$$ (the other choice will give the same result).
Here $$\lambda=-1,\;\mu=1$$ and
$$3\lambda\hat{d}=3(-1)\Bigl(-\tfrac23,-\tfrac13,\tfrac23\Bigr)=(2,1,-2),$$
$$\mu\vec{c}=1\,(0,-1,1)=(0,-1,1).$$

Therefore
$$3\lambda\hat{d}+\mu\vec{c}=(2,1,-2)+(0,-1,1)=(2,0,-1).$$

The square of its magnitude is
$$\lvert(2,0,-1)\rvert^{2}=2^{2}+0^{2}+(-1)^{2}=4+1=5.$$

Repeating with $$s=-1$$ also produces the vector $$(2,0,-1)$$, confirming the result is independent of the sign choice.

Final Answer:
$$\bigl|\,3\lambda\hat{d}+\mu\vec{c}\bigr|^{2}=5.$$

Given $$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$$, $$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$$, $$\vec{c} = 2\hat{i} - \hat{j} + 2\hat{k}$$.

We need $$\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$$, which gives $$(\vec{b} - \vec{c}) \times \vec{d} = \vec{0}$$.

$$\vec{b} - \vec{c} = (3-2)\hat{i} + (-3+1)\hat{j} + (3-2)\hat{k} = \hat{i} - 2\hat{j} + \hat{k}$$

Since $$(\vec{b} - \vec{c}) \times \vec{d} = \vec{0}$$, $$\vec{d}$$ is parallel to $$\vec{b} - \vec{c} = \hat{i} - 2\hat{j} + \hat{k}$$.

So $$\vec{d} = t(\hat{i} - 2\hat{j} + \hat{k})$$ for some scalar $$t$$.

Using $$\vec{a} \cdot \vec{d} = 4$$: $$t(1 - 4 + 1) = 4$$, so $$t(-2) = 4$$, giving $$t = -2$$.

Therefore, $$\vec{d} = -2\hat{i} + 4\hat{j} - 2\hat{k}$$.

Now, $$\vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 4 & -2 \end{vmatrix}$$

$$= \hat{i}(-4 - 4) - \hat{j}(-2 + 2) + \hat{k}(4 + 4) = -8\hat{i} + 0\hat{j} + 8\hat{k}$$

$$|\vec{a} \times \vec{d}|^2 = 64 + 0 + 64 = 128$$

Hence, the correct answer is 128.

Given vectors:

$$\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}$$

$$\overrightarrow{b} = 2\hat{i} + 2\hat{j} + \hat{k}$$

First, compute $$\overrightarrow{d} = \overrightarrow{a} \times \overrightarrow{b}$$.

The cross product is:

$$\overrightarrow{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 1 \cdot 2) - \hat{j}(1 \cdot 1 - 1 \cdot 2) + \hat{k}(1 \cdot 2 - 1 \cdot 2) = \hat{i}(-1) - \hat{j}(-1) + \hat{k}(0) = -\hat{i} + \hat{j}$$

So, $$\overrightarrow{d} = -\hat{i} + \hat{j}$$.

Let $$\overrightarrow{c} = x\hat{i} + y\hat{j} + z\hat{k}$$. The given conditions are:

1. $$\overrightarrow{a} \cdot \overrightarrow{c} = |\overrightarrow{c}|$$

$$\overrightarrow{a} \cdot \overrightarrow{c} = x + y + z$$

$$|\overrightarrow{c}| = \sqrt{x^2 + y^2 + z^2}$$

So,

$$x + y + z = \sqrt{x^2 + y^2 + z^2} \quad \text{(1)}$$

2. $$|\overrightarrow{c} - 2\overrightarrow{a}|^2 = 8$$

$$2\overrightarrow{a} = 2\hat{i} + 2\hat{j} + 2\hat{k}$$

$$\overrightarrow{c} - 2\overrightarrow{a} = (x - 2)\hat{i} + (y - 2)\hat{j} + (z - 2)\hat{k}$$

So,

$$(x - 2)^2 + (y - 2)^2 + (z - 2)^2 = 8 \quad \text{(2)}$$

3. Angle between $$\overrightarrow{d}$$ and $$\overrightarrow{c}$$ is $$\frac{\pi}{4}$$.

$$\overrightarrow{d} \cdot \overrightarrow{c} = (-1)(x) + (1)(y) + (0)(z) = -x + y$$

$$|\overrightarrow{d}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$$

$$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

So,

$$\overrightarrow{d} \cdot \overrightarrow{c} = |\overrightarrow{d}| |\overrightarrow{c}| \cos \frac{\pi}{4} \implies -x + y = \sqrt{2} \cdot |\overrightarrow{c}| \cdot \frac{\sqrt{2}}{2} = |\overrightarrow{c}|$$

Thus,

$$-x + y = |\overrightarrow{c}| \quad \text{(3)}$$

From equations (1) and (3), let $$|\overrightarrow{c}| = k$$:

$$x + y + z = k \quad \text{(1)}$$

$$-x + y = k \quad \text{(3)}$$

Subtract equation (3) from equation (1):

$$(x + y + z) - (-x + y) = k - k \implies 2x + z = 0 \quad \text{(4)}$$

So, $$z = -2x$$.

From equation (3):

$$y = k + x \quad \text{(5)}$$

Substitute $$z = -2x$$ and $$y = k + x$$ into equation (1):

$$x + (k + x) + (-2x) = k \implies k = k$$ (always true).

Now, $$k = |\overrightarrow{c}| = \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2 + (k + x)^2 + (-2x)^2}$$:

$$k^2 = x^2 + (k + x)^2 + (-2x)^2 = x^2 + k^2 + 2kx + x^2 + 4x^2 = k^2 + 6x^2 + 2kx$$

$$0 = 6x^2 + 2kx \implies 2x(3x + k) = 0$$

So, $$x = 0$$ or $$k = -3x$$.

**Case 1: $$x = 0$$**

From equation (4): $$z = -2(0) = 0$$.

From equation (5): $$y = k + 0 = k$$.

From equation (1): $$0 + k + 0 = k \implies k = k$$ (true).

Equation (2):

$$(0 - 2)^2 + (k - 2)^2 + (0 - 2)^2 = 4 + (k - 2)^2 + 4 = (k - 2)^2 + 8 = 8$$

$$(k - 2)^2 = 0 \implies k = 2$$

Then $$y = 2$$, so $$\overrightarrow{c} = 2\hat{j}$$.

**Case 2: $$k = -3x$$**

Since $$k \geq 0$$, $$-3x \geq 0 \implies x \leq 0$$.

From equation (5): $$y = -3x + x = -2x$$.

From equation (4): $$z = -2x$$.

Equation (2):

$$(x - 2)^2 + (-2x - 2)^2 + (-2x - 2)^2 = (x - 2)^2 + 2(-2(x + 1))^2 = (x - 2)^2 + 8(x + 1)^2 = 8$$

$$(x - 2)^2 + 8(x + 1)^2 = 8$$

$$x^2 - 4x + 4 + 8(x^2 + 2x + 1) = 8 \implies x^2 - 4x + 4 + 8x^2 + 16x + 8 = 8 \implies 9x^2 + 12x + 12 = 8$$

$$9x^2 + 12x + 4 = 0$$

Discriminant: $$12^2 - 4 \cdot 9 \cdot 4 = 144 - 144 = 0$$

$$x = \frac{-12}{18} = -\frac{2}{3}$$

Then $$k = -3 \left(-\frac{2}{3}\right) = 2$$, $$y = -2 \left(-\frac{2}{3}\right) = \frac{4}{3}$$, $$z = -2 \left(-\frac{2}{3}\right) = \frac{4}{3}$$, so $$\overrightarrow{c} = -\frac{2}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{4}{3}\hat{k}$$.

Now compute $$|10 - 3\overrightarrow{b} \cdot \overrightarrow{c}| + |\overrightarrow{d} \times \overrightarrow{c}|^2$$ for both cases.

**For $$\overrightarrow{c} = 2\hat{j}$$:**

$$\overrightarrow{b} \cdot \overrightarrow{c} = (2)(0) + (2)(2) + (1)(0) = 4$$

$$3\overrightarrow{b} \cdot \overrightarrow{c} = 12$$

$$10 - 12 = -2 \implies | -2 | = 2$$

$$\overrightarrow{d} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 0 & 2 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0) + \hat{k}(-2) = -2\hat{k}$$

$$|\overrightarrow{d} \times \overrightarrow{c}|^2 = (-2)^2 = 4$$

Sum: $$2 + 4 = 6$$.

**For $$\overrightarrow{c} = -\frac{2}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{4}{3}\hat{k}$$:**

$$\overrightarrow{b} \cdot \overrightarrow{c} = (2)\left(-\frac{2}{3}\right) + (2)\left(\frac{4}{3}\right) + (1)\left(\frac{4}{3}\right) = -\frac{4}{3} + \frac{8}{3} + \frac{4}{3} = \frac{8}{3}$$

$$3\overrightarrow{b} \cdot \overrightarrow{c} = 8$$

$$10 - 8 = 2 \implies |2| = 2$$

$$\overrightarrow{d} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ -\frac{2}{3} & \frac{4}{3} & \frac{4}{3} \end{vmatrix} = \hat{i}\left(1 \cdot \frac{4}{3} - 0 \cdot \frac{4}{3}\right) - \hat{j}\left(-1 \cdot \frac{4}{3} - 0 \cdot \left(-\frac{2}{3}\right)\right) + \hat{k}\left(-1 \cdot \frac{4}{3} - 1 \cdot \left(-\frac{2}{3}\right)\right) = \frac{4}{3}\hat{i} + \frac{4}{3}\hat{j} - \frac{2}{3}\hat{k}$$

$$|\overrightarrow{d} \times \overrightarrow{c}|^2 = \left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 = \frac{16}{9} + \frac{16}{9} + \frac{4}{9} = \frac{36}{9} = 4$$

Sum: $$2 + 4 = 6$$.

In both cases, the expression evaluates to 6.

Thus, $$|10 - 3\overrightarrow{b} \cdot \overrightarrow{c}| + |\overrightarrow{d} \times \overrightarrow{c}|^2 = 6$$.

**Final Answer:** $$6$$

Given the side vectors of triangle $$ABC$$ be:

$$\vec{AB} = 2\hat{i} - \hat{j} + \hat{k}$$

$$\vec{BC} = \hat{i} - 3\hat{j} - 5\hat{k}$$

$$\vec{AC} = \vec{AB} + \vec{BC} = 3\hat{i} - 4\hat{j} - 4\hat{k}$$

---

Calculating the squares of the lengths of the sides of the triangle:

$$c^2 = |\vec{AB}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6$$

$$a^2 = |\vec{BC}|^2 = 1^2 + (-3)^2 + (-5)^2 = 1 + 9 + 25 = 35$$

$$b^2 = |\vec{AC}|^2 = 3^2 + (-4)^2 + (-4)^2 = 9 + 16 + 16 = 41$$

The sum of the squares of the sides is:

$$a^2 + b^2 + c^2 = 35 + 41 + 6 = 82$$

---

Using the centroid property for any triangle $$ABC$$ with centroid $$G$$:

$$|\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2 = \frac{1}{3}\left(a^2 + b^2 + c^2\right)$$

Substituting the value of the sum of the squares of the sides:

$$|\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2 = \frac{82}{3}$$

---

Multiplying the expression by 6 to find the required value:

$$6\left(|\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2\right) = 6 \times \frac{82}{3} = 2 \times 82 = 164$$

Therefore, the final value is equal to 164.

Let the position vectors of points A, B and C be given by $$\overrightarrow{OA}=\sqrt{3}\,\hat{i}+\hat{j}$$, $$\overrightarrow{OB}=\hat{i}+\sqrt{3}\,\hat{j}$$ and $$\overrightarrow{OC}=a\,\hat{i}+(1-a)\,\hat{j}$$ respectively.

We first find the equation of the internal angle bisector of the angle between $$\overrightarrow{OA}$$ and $$\overrightarrow{OB}$$.

Formula for internal bisector direction: if $$\mathbf{v}_1$$ and $$\mathbf{v}_2$$ are two vectors, then a direction vector along the internal bisector is $$\mathbf{u}=\frac{\mathbf{v}_1}{\|\mathbf{v}_1\|}+\frac{\mathbf{v}_2}{\|\mathbf{v}_2\|}$$.

Here, $$\|\overrightarrow{OA}\|=\sqrt{(\sqrt{3})^2+1^2}=\sqrt{3+1}=2$$ and $$\|\overrightarrow{OB}\|=\sqrt{1^2+(\sqrt{3})^2}=2$$.

Thus $$\mathbf{u} =\frac{1}{2}(\sqrt{3},1)+\frac{1}{2}(1,\sqrt{3}) =\Bigl(\frac{\sqrt{3}+1}{2},\frac{1+\sqrt{3}}{2}\Bigr) \;\Longrightarrow\; \text{direction }(d_x,d_y)=(\sqrt{3}+1,\;1+\sqrt{3}).$$

Since $$\sqrt{3}+1=1+\sqrt{3}$$, the slope of this line is $$m=\frac{1+\sqrt{3}}{\sqrt{3}+1}=1.$$ Hence the internal bisector is the line $$y-x=0\quad\text{or}\quad x-y=0.$$

Point C has coordinates $$(a,\;1-a)$$. The distance of $$(a,1-a)$$ from the line $$x-y=0$$ is given by the formula $$\text{Distance}=\frac{\bigl|a-(1-a)\bigr|}{\sqrt{1^2+(-1)^2}} =\frac{|2a-1|}{\sqrt{2}}\quad-(1)$$.

We are given that this distance equals $$\dfrac{9}{\sqrt{2}}$$. Equating with $$(1)$$ we get: $$\frac{|2a-1|}{\sqrt{2}}=\frac{9}{\sqrt{2}} \;\Longrightarrow\;|2a-1|=9.$$ This gives two cases:

Case 1: $$2a-1=9\quad\Longrightarrow\quad2a=10\quad\Longrightarrow\quad a=5.$$

Case 2: $$2a-1=-9\quad\Longrightarrow\quad2a=-8\quad\Longrightarrow\quad a=-4.$$

Thus the possible values of $$a$$ are $$5$$ and $$-4$$. Their sum is $$5+(-4)=1.$$

Final Answer: The sum of all possible values of $$a$$ is $$1$$ (Option C).

The position vectors of vertices A, B, and C are given as: $$\overrightarrow{A} = \widehat{i} + 2\widehat{j} + \widehat{k},\quad \overrightarrow{B} = \widehat{i} + 3\widehat{j} - 2\widehat{k},\quad \overrightarrow{C} = 2\widehat{i} + \widehat{j} - \widehat{k}.$$

The median from vertex A to side BC is found by determining the midpoint M of BC: $$\overrightarrow{M} = \frac{\overrightarrow{B} + \overrightarrow{C}}{2} = \frac{(\widehat{i} + 3\widehat{j} - 2\widehat{k}) + (2\widehat{i} + \widehat{j} - \widehat{k})}{2} = \frac{3\widehat{i} + 4\widehat{j} - 3\widehat{k}}{2} = \frac{3}{2}\widehat{i} + 2\widehat{j} - \frac{3}{2}\widehat{k}.$$ The vector $$\overrightarrow{AM}$$ is $$\overrightarrow{AM} = \overrightarrow{M} - \overrightarrow{A} = \Bigl(\tfrac{3}{2}\widehat{i} + 2\widehat{j} - \tfrac{3}{2}\widehat{k}\Bigr) - (\widehat{i} + 2\widehat{j} + \widehat{k}) = \tfrac{1}{2}\widehat{i} - \tfrac{5}{2}\widehat{k}.$$

The parametric equation of the median AM is $$\overrightarrow{P}(t) = \overrightarrow{A} + t\,\overrightarrow{AM} = (\widehat{i} + 2\widehat{j} + \widehat{k}) + t\Bigl(\tfrac{1}{2}\widehat{i} - \tfrac{5}{2}\widehat{k}\Bigr) = \Bigl(1 + \tfrac{t}{2}\Bigr)\widehat{i} + 2\widehat{j} + \Bigl(1 - \tfrac{5t}{2}\Bigr)\widehat{k}.$$

The altitude from D to the plane ABC is perpendicular to the plane. The normal vector to plane ABC is $$\overrightarrow{AB}\times\overrightarrow{AC}$$, where $$\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A} = (\widehat{i} + 3\widehat{j} - 2\widehat{k}) - (\widehat{i} + 2\widehat{j} + \widehat{k}) = \widehat{j} - 3\widehat{k},$$ $$\overrightarrow{AC} = \overrightarrow{C} - \overrightarrow{A} = (2\widehat{i} + \widehat{j} - \widehat{k}) - (\widehat{i} + 2\widehat{j} + \widehat{k}) = \widehat{i} - \widehat{j} - 2\widehat{k}.$$ Their cross product is $$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 0 & 1 & -3 \\ 1 & -1 & -2 \end{vmatrix} = \widehat{i}(-2 - 3) - \widehat{j}(0 + 3) + \widehat{k}(0 - 1) = -5\widehat{i} - 3\widehat{j} - \widehat{k}.$$ Thus the direction of the altitude is parallel to $$\overrightarrow{N} = -5\widehat{i} - 3\widehat{j} - \widehat{k}$$, and its parametric equation from D is $$\overrightarrow{Q}(s) = \overrightarrow{D} + s\,\overrightarrow{N} = (x\widehat{i} + y\widehat{j} + z\widehat{k}) + s(-5\widehat{i} - 3\widehat{j} - \widehat{k}) = (x - 5s)\widehat{i} + (y - 3s)\widehat{j} + (z - s)\widehat{k}.$$

The intersection point of the altitude and the median satisfies $$(x - 5s)\widehat{i} + (y - 3s)\widehat{j} + (z - s)\widehat{k} = \Bigl(1 + \tfrac{t}{2}\Bigr)\widehat{i} + 2\widehat{j} + \Bigl(1 - \tfrac{5t}{2}\Bigr)\widehat{k},$$ so equating components gives $$x - 5s = 1 + \tfrac{t}{2},\quad y - 3s = 2,\quad z - s = 1 - \tfrac{5t}{2}.$$ From $$y - 3s = 2$$ we get $$y = 2 + 3s$$. Defining $$u = x - 1,\ v = y - 2,\ w = z - 1$$, then $$v = 3s$$. From $$x - 5s = 1 + t/2$$ we have $$u - 5s = t/2\implies t = 2(u - 5s)$$, and from $$z - s = 1 - 5t/2$$ we obtain $$w - s = -5t/2$$. Substituting $$t = 2(u - 5s)$$ yields $$w - s = -5(u - 5s)\implies w = -5u + 26s.$$

The volume of the tetrahedron is $$\tfrac{\sqrt{805}}{6\sqrt{2}}$$, so the scalar triple product satisfies $$\bigl|\overrightarrow{AD}\cdot(\overrightarrow{AB}\times\overrightarrow{AC})\bigr| = \bigl|-5u - 3v - w\bigr| = \frac{\sqrt{805}}{\sqrt{2}} = \sqrt{\tfrac{805}{2}}.$$ Substituting $$v = 3s$$ and $$w = -5u + 26s$$ gives $$S = -5u - 9s + 5u - 26s = -35s,\quad |S| = 35|s| = \sqrt{\tfrac{805}{2}},$$ so $$1225s^2 = \frac{805}{2}\implies s^2 = \frac{23}{70}.$$

The distance $$AD$$ is $$\tfrac{\sqrt{110}}{3}$$, hence $$|\overrightarrow{AD}|^2 = u^2 + v^2 + w^2 = u^2 + 9s^2 + (-5u + 26s)^2 = \frac{110}{9}.$$ Expanding yields $$26u^2 - 260us + 685s^2 = \frac{110}{9},$$ and substituting $$s^2 = \frac{23}{70}$$ gives the quadratic $$252u^2 - 2520su + 2063 = 0.$$

Solving for $$u$$: $$u = \frac{2520s \pm \sqrt{(2520s)^2 - 4\cdot252\cdot2063}}{504} = \frac{2520s \pm 84}{504} = 5s \pm \tfrac{1}{6}.$$ Thus $$u = 5s + \tfrac{1}{6}$$ or $$u = 5s - \tfrac{1}{6}$$, giving $$t = \tfrac{1}{3}$$ or $$t = -\tfrac{1}{3}$$ respectively.

The intersection point $$P$$ on the median is then: for $$t = \tfrac{1}{3}$$, $$\overrightarrow{P} = \tfrac{1}{6}(7\widehat{i} + 12\widehat{j} + \widehat{k});$$ for $$t = -\tfrac{1}{3}$$, $$\overrightarrow{P} = \tfrac{1}{6}(5\widehat{i} + 12\widehat{j} + 11\widehat{k}).$$ Comparing with the options shows that Option D, $$\tfrac{1}{6}(7\widehat{i} + 12\widehat{j} + \widehat{k})$$, matches the case $$t = \tfrac{1}{3}$$. Verification with $$AD$$ length and volume confirms consistency.

Final Answer: D. $$\frac{1}{6}\left(7\widehat{i}+12\widehat{j}+\widehat{k}\right)$$

Line L: $$\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2} = t$$

Parametric form: $$(1+t, -1-t, 2+2t)$$
Direction of L: $$\vec{d} = (1,-1,2)$$
Vector from point on L to (1,2,2): $$(1-(1+t), 2-(-1-t), 2-(2+2t)) = (-t, 3+t, -2t)$$
For perpendicularity: $$(-t)(1) + (3+t)(-1) + (-2t)(2) = 0$$
$$-t - 3 - t - 4t = 0$$
$$t = -\frac{1}{2}$$

So $$P = \left(\frac{1}{2}, -\frac{1}{2}, 1\right)$$.

The second line: $$\vec{r} = (-1,1,-2) + \lambda(1,-1,1)$$, i.e., $$(-1+\lambda, 1-\lambda, -2+\lambda)$$
$$-1+\lambda = 1+t \Rightarrow \lambda = 2+t \quad ...(1)$$
$$1-\lambda = -1-t \Rightarrow \lambda = 2+t \quad ...(2)$$ ✓
$$-2+\lambda = 2+2t \quad ...(3)$$

From (1): $$\lambda = 2+t$$. Substituting in (3):
$$t = -2$$
$$\lambda = 0$$

So $$Q = (-1, 1, -2)$$.
$$PQ = Q - P = \left(-1-\frac{1}{2}, 1+\frac{1}{2}, -2-1\right) = \left(-\frac{3}{2}, \frac{3}{2}, -3\right)$$
$$(PQ)^2 = \frac{9}{4} + \frac{9}{4} + 9 = \frac{9}{4} + \frac{9}{4} + \frac{36}{4} = \frac{54}{4} = \frac{27}{2}$$
$$2(PQ)^2 = 2 \times \frac{27}{2} = 27$$

We are given $$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$$ and $$\vec{b} = 2\hat{i} + 7\hat{j}+ 3\hat{k}$$, and the lines:

$$L_1: \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda\vec{a}$$

$$L_2: \vec{r} = (\hat{j} + \hat{k}) + \mu\vec{b}$$

Line $$L_3$$ passes through the intersection of $$L_1$$ and $$L_2$$ and is parallel to $$\vec{a} + \vec{b}$$.

Points on $$L_1$$ are given by $$(-1+\lambda,\;2+2\lambda,\;1+\lambda)$$ and points on $$L_2$$ by $$(2\mu,\;1+7\mu,\;1+3\mu)$$. Setting these equal gives:

$$-1+\lambda = 2\mu\,,\quad 2+2\lambda = 1+7\mu\,,\quad 1+\lambda = 1+3\mu$$

From the third equation $$\lambda = 3\mu$$, and substituting into the first yields $$-1+3\mu = 2\mu\implies \mu = 1$$ and hence $$\lambda = 3$$. 

Verification in the second equation shows $$2+6=8$$ and $$1+7=8$$. The intersection point is therefore $$(-1+3,\;2+6,\;1+3) = (2,8,4)\,.$$

The direction of $$L_3$$ is the vector sum:

$$\vec{a} + \vec{b} = (1+2)\hat{i} + (2+7)\hat{j} + (1+3)\hat{k} = 3\hat{i} + 9\hat{j} + 4\hat{k}$$

The equation of $$L_3$$ can then be written in vector form as:

$$\vec{r} = (2\hat{i} + 8\hat{j} + 4\hat{k}) + t\,(3\hat{i} + 9\hat{j} + 4\hat{k})$$

with parametric form $$(2+3t,\;8+9t,\;4+4t)\,.$$

Checking which point lies on $$L_3$$, for $$(8,26,12)$$ we have $$2+3t=8\implies t=2$$, and then $$8+18=26$$ and $$4+8=12$$, so the choice is consistent. Thus the correct answer is Option 3: $$(8,26,12)\,.$$

Given vectors are $$\overrightarrow{a} = 2\hat{i} - \hat{j} + 3\hat{k}$$ and $$\overrightarrow{b} = 3\hat{i} - 5\hat{j} + \hat{k}$$. The conditions are:

1. $$\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{b}$$
2. $$(\overrightarrow{a} + \overrightarrow{c}) \cdot (\overrightarrow{b} \cdot \overrightarrow{c}) = 168$$, but this appears to have a typo. Given the context and solution, it is interpreted as $$(\overrightarrow{a} + \overrightarrow{c}) \cdot (\overrightarrow{b} + \overrightarrow{c}) = 168$$.

From the first condition:

$$\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{b}$$
Since $$\overrightarrow{c} \times \overrightarrow{b} = - \overrightarrow{b} \times \overrightarrow{c}$$, we have:
$$\overrightarrow{a} \times \overrightarrow{c} = - \overrightarrow{b} \times \overrightarrow{c}$$
$$\overrightarrow{a} \times \overrightarrow{c} + \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{0}$$
$$(\overrightarrow{a} + \overrightarrow{b}) \times \overrightarrow{c} = \overrightarrow{0}$$
Thus, $$\overrightarrow{c}$$ is parallel to $$\overrightarrow{a} + \overrightarrow{b}$$.

Compute $$\overrightarrow{a} + \overrightarrow{b}$$:
$$\overrightarrow{a} + \overrightarrow{b} = (2 + 3)\hat{i} + (-1 - 5)\hat{j} + (3 + 1)\hat{k} = 5\hat{i} - 6\hat{j} + 4\hat{k}$$
Let $$\overrightarrow{d} = 5\hat{i} - 6\hat{j} + 4\hat{k}$$, so $$\overrightarrow{c} = k \overrightarrow{d} = k(5\hat{i} - 6\hat{j} + 4\hat{k})$$ for some scalar $$k$$.

Now, the second condition is $$(\overrightarrow{a} + \overrightarrow{c}) \cdot (\overrightarrow{b} + \overrightarrow{c}) = 168$$.
Substitute $$\overrightarrow{c} = k(5\hat{i} - 6\hat{j} + 4\hat{k})$$:
$$\overrightarrow{a} + \overrightarrow{c} = (2 + 5k)\hat{i} + (-1 - 6k)\hat{j} + (3 + 4k)\hat{k}$$
$$\overrightarrow{b} + \overrightarrow{c} = (3 + 5k)\hat{i} + (-5 - 6k)\hat{j} + (1 + 4k)\hat{k}$$

Dot product:
$$(\overrightarrow{a} + \overrightarrow{c}) \cdot (\overrightarrow{b} + \overrightarrow{c}) = (2 + 5k)(3 + 5k) + (-1 - 6k)(-5 - 6k) + (3 + 4k)(1 + 4k)$$
Expand each term:
$$(2 + 5k)(3 + 5k) = 2 \cdot 3 + 2 \cdot 5k + 5k \cdot 3 + 5k \cdot 5k = 6 + 10k + 15k + 25k^2 = 6 + 25k + 25k^2$$
$$(-1 - 6k)(-5 - 6k) = (-1)(-5) + (-1)(-6k) + (-6k)(-5) + (-6k)(-6k) = 5 + 6k + 30k + 36k^2 = 5 + 36k + 36k^2$$
$$(3 + 4k)(1 + 4k) = 3 \cdot 1 + 3 \cdot 4k + 4k \cdot 1 + 4k \cdot 4k = 3 + 12k + 4k + 16k^2 = 3 + 16k + 16k^2$$

Sum the expressions:
$$(6 + 5 + 3) + (25k + 36k + 16k) + (25k^2 + 36k^2 + 16k^2) = 14 + 77k + 77k^2$$

Set equal to 168:
$$77k^2 + 77k + 14 = 168$$
$$77k^2 + 77k - 154 = 0$$
Divide by 77:
$$k^2 + k - 2 = 0$$
Solve the quadratic equation:
Discriminant $$D = 1^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9$$
$$k = \frac{-1 \pm \sqrt{9}}{2} = \frac{-1 \pm 3}{2}$$
So, $$k = \frac{-1 + 3}{2} = 1$$ or $$k = \frac{-1 - 3}{2} = -2$$

Now, $$|\overrightarrow{c}|^2 = k^2 |\overrightarrow{d}|^2$$
$$|\overrightarrow{d}|^2 = 5^2 + (-6)^2 + 4^2 = 25 + 36 + 16 = 77$$
For $$k = 1$$, $$|\overrightarrow{c}|^2 = (1)^2 \cdot 77 = 77$$
For $$k = -2$$, $$|\overrightarrow{c}|^2 = (-2)^2 \cdot 77 = 4 \cdot 77 = 308$$
The maximum value is 308.

Verify the second condition for both values:
For $$k = 1$$, $$\overrightarrow{c} = 5\hat{i} - 6\hat{j} + 4\hat{k}$$
$$\overrightarrow{a} + \overrightarrow{c} = (2 + 5)\hat{i} + (-1 - 6)\hat{j} + (3 + 4)\hat{k} = 7\hat{i} - 7\hat{j} + 7\hat{k}$$
$$\overrightarrow{b} + \overrightarrow{c} = (3 + 5)\hat{i} + (-5 - 6)\hat{j} + (1 + 4)\hat{k} = 8\hat{i} - 11\hat{j} + 5\hat{k}$$
Dot product: $$7 \cdot 8 + (-7) \cdot (-11) + 7 \cdot 5 = 56 + 77 + 35 = 168$$
For $$k = -2$$, $$\overrightarrow{c} = -10\hat{i} + 12\hat{j} - 8\hat{k}$$
$$\overrightarrow{a} + \overrightarrow{c} = (2 - 10)\hat{i} + (-1 + 12)\hat{j} + (3 - 8)\hat{k} = -8\hat{i} + 11\hat{j} - 5\hat{k}$$
$$\overrightarrow{b} + \overrightarrow{c} = (3 - 10)\hat{i} + (-5 + 12)\hat{j} + (1 - 8)\hat{k} = -7\hat{i} + 7\hat{j} - 7\hat{k}$$
Dot product: $$(-8) \cdot (-7) + 11 \cdot 7 + (-5) \cdot (-7) = 56 + 77 + 35 = 168$$

Both satisfy the second condition, and the first condition is satisfied since $$\overrightarrow{c}$$ is parallel to $$\overrightarrow{a} + \overrightarrow{b}$$ for both $$k$$. The maximum value of $$|\overrightarrow{c}|^2$$ is 308.

Final Answer: 308

We want a unit vector $$\hat{c}$$ which

• lies in the plane of $$\vec{a}$$ and $$\vec{b}$$ (so it is a linear combination of them), and
• is perpendicular to $$\vec{a}$$.

Write $$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$$ and $$\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$$.

A convenient way to obtain a vector lying in the required plane and orthogonal to $$\vec{a}$$ is to remove from $$\vec{b}$$ its component along $$\vec{a}$$. The result is automatically in the span of $$\vec{a},\vec{b}$$ and perpendicular to $$\vec{a}$$.

First compute the projection of $$\vec{b}$$ on $$\vec{a}$$.
Formula: projection of $$\vec{b}$$ on $$\vec{a}$$ is $$\displaystyle\frac{\vec{a}\cdot\vec{b}}{\vec{a}\cdot\vec{a}}\;\vec{a}$$.

Dot products:
$$\vec{a}\cdot\vec{b} = 1\cdot2 + 2\cdot1 + 1\cdot(-1) = 2 + 2 - 1 = 3$$
$$\vec{a}\cdot\vec{a} = 1^2 + 2^2 + 1^2 = 1 + 4 + 1 = 6$$

Hence the projection is
$$\frac{3}{6}\,\vec{a} = \frac{1}{2}\,( \hat{i} + 2\hat{j} + \hat{k}) = 0.5\,\hat{i} + 1\,\hat{j} + 0.5\,\hat{k}$$.

Subtract this from $$\vec{b}$$ to get the component of $$\vec{b}$$ perpendicular to $$\vec{a}$$:

$$\vec{d} = \vec{b} - \text{proj}_{\vec{a}}\vec{b}$$

$$= (2,\,1,\,-1) - (0.5,\,1,\,0.5)$$

$$= (1.5,\,0,\,-1.5)$$.

Multiply by $$2$$ to clear the decimal:

$$\vec{d} = (3,\,0,\,-3) = 3(1,\,0,\,-1).$$

Any non-zero scalar multiple of $$\vec{d}$$ is still in the same direction, so we choose the simple vector
$$\vec{m} = (1,\,0,\,-1) = \hat{i} - \hat{k}$$.

Check orthogonality with $$\vec{a}$$:
$$(\hat{i} - \hat{k})\cdot(\hat{i} + 2\hat{j} + \hat{k}) = 1 - 1 = 0,$$
so it is indeed perpendicular to $$\vec{a}$$.

Now normalise to get a unit vector:

Magnitude of $$\vec{m}$$:
$$|\vec{m}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$$.

Hence

$$\hat{c} = \frac{\vec{m}}{|\vec{m}|} = \frac{1}{\sqrt{2}}(\hat{i} - \hat{k}).$$

The negative of a unit vector is also a valid unit vector, so $$-\hat{c} = \dfrac{1}{\sqrt{2}}(-\hat{i} + \hat{k})$$ is equally acceptable.

Comparing with the options, Option D gives $$\dfrac{1}{\sqrt{2}}(-\hat{i} + \hat{k})$$, which matches (up to sign).

Therefore the required vector is given by Option D.

Write the given vectors in component form:
$$\vec{p}=2\hat i+\hat j+3\hat k \; \rightarrow \; (2,1,3), \qquad \vec{q}=\hat i-\hat j+\hat k \; \rightarrow \; (1,-1,1).$$

First compute the three auxiliary vectors that appear on the right-hand side.

1. Vector $$2\vec p+\vec q$$
$$2\vec p=(4,2,6) \quad\Longrightarrow\quad 2\vec p+\vec q=(4+1,\,2+(-1),\,6+1)=(5,1,7).$$

2. Vector $$\vec p-2\vec q$$
$$2\vec q=(2,-2,2) \quad\Longrightarrow\quad \vec p-2\vec q=(2-2,\,1-(-2),\,3-2)=(0,3,1).$$

3. Vector $$\vec p\times\vec q$$
Using the determinant rule:
$$ \vec p\times\vec q= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 1 & 3\\ 1 & -1 & 1 \end{vmatrix} = \hat i(1\cdot1-3\cdot(-1))-\hat j(2\cdot1-3\cdot1)+\hat k(2\cdot(-1)-1\cdot1) =4\hat i+\hat j-3\hat k. $$
Thus $$\vec p\times\vec q=(4,1,-3).$$

Denote
$$\vec a_1=(5,1,7),\qquad \vec a_2=(0,3,1),\qquad \vec a_3=(4,1,-3).$$
The equation in component form becomes

$$\alpha\vec a_1+\beta\vec a_2+\gamma\vec a_3=(15,10,6).$$

Equate the $$\hat i, \hat j, \hat k$$ components:

$$5\alpha+4\gamma=15 \quad-(1)$$
$$\alpha+3\beta+\gamma=10 \quad-(2)$$
$$7\alpha+\beta-3\gamma=6 \quad-(3)$$

Solve sequentially. From $$(1):$$
$$\alpha=\frac{15-4\gamma}{5}=3-0.8\gamma.$$

Substitute this $$\alpha$$ into $$(2):$$
$$(3-0.8\gamma)+3\beta+\gamma=10 \;\Longrightarrow\; 3\beta=7-0.2\gamma,$$
so $$\beta=\dfrac{7-0.2\gamma}{3}.$$

Now use $$(3):$$
$$7\alpha+\beta-3\gamma=6.$$ Insert $$\alpha=3-0.8\gamma$$ and the above $$\beta$$:

$$7(3-0.8\gamma)+\beta-3\gamma=6$$ $$\Longrightarrow 21-5.6\gamma+\beta-3\gamma=6$$ $$\Longrightarrow \beta+21-8.6\gamma=6$$ $$\Longrightarrow \beta=-15+8.6\gamma.$$

Equate the two expressions for $$\beta$$:

$$\frac{7-0.2\gamma}{3}=-15+8.6\gamma.$$ Multiply by 3: $$7-0.2\gamma=-45+25.8\gamma$$ $$\Longrightarrow 52=26\gamma$$ $$\Longrightarrow \gamma=2.$$

Hence, the required value is $$\boxed{2}$$.

The components of the three position vectors are
$$\overrightarrow{OP} = \left( \frac{\alpha-1}{\alpha},\, 1,\, 1 \right),\; \overrightarrow{OQ} = \left( 1,\, \frac{\beta-1}{\beta},\, 1 \right),\; \overrightarrow{OR} = \left( 1,\, 1,\, \frac12 \right).$$

For any three vectors $$\mathbf{a},\mathbf{b},\mathbf{c}$$ originating from one point, the scalar triple product is the determinant of the matrix formed by their components:
$$(\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}= \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix}.$$

Using $$\mathbf{a}=\overrightarrow{OP},\;\mathbf{b}=\overrightarrow{OQ},\;\mathbf{c}=\overrightarrow{OR}$$ we get

$$ \begin{vmatrix} \dfrac{\alpha-1}{\alpha} & 1 & 1 \\[4pt] 1 & \dfrac{\beta-1}{\beta} & 1 \\[4pt] 1 & 1 & \dfrac12 \end{vmatrix}=0.$$ Denote $$A = 1-\frac1{\alpha},\qquad B = 1-\frac1{\beta};$$ then the determinant becomes

$$ \begin{vmatrix} A & 1 & 1 \\ 1 & B & 1 \\ 1 & 1 & \dfrac12 \end{vmatrix}=0.$$ Expanding along the first row:

$$ A\!\left(B\cdot\frac12-1\cdot1\right) -\;1\!\left(1\cdot\frac12-1\cdot1\right) +\;1\!\left(1\cdot1-B\cdot1\right)=0. $$

Simplifying each term:
$$A\!\left(\frac{B}{2}-1\right)+\frac12+(1-B)=0,$$ so
$$A\!\left(\frac{B}{2}-1\right)+\frac32-B=0.$$

Substitute $$A=1-\dfrac1\alpha,\;B=1-\dfrac1\beta$$:

$$ \left(1-\frac1\alpha\right)\!\left(\frac12-\frac1{2\beta}-1\right)+\frac32-\left(1-\frac1\beta\right)=0. $$ First bracket: $$\frac12-\frac1{2\beta}-1=-\frac12-\frac1{2\beta}.$$

Hence
$$-\left(1-\frac1\alpha\right)\!\left(\frac12+\frac1{2\beta}\right)+\frac12+\frac1\beta=0.$$

Multiply throughout by $$2\beta$$ to clear denominators:

$$-\left(1-\frac1\alpha\right)(\beta+1)+\beta+2=0.$$

Expand:
$$-(\beta+1)+\frac{\beta+1}{\alpha}+\beta+2=0.$$

Combining like terms:
$$1+\frac{\beta+1}{\alpha}=0\;\;\Longrightarrow\;\;\frac{\beta+1}{\alpha}=-1 \;\;\Longrightarrow\;\;\beta=-\alpha-1.$$

The point $$(\alpha,\beta,2)$$ lies on the plane $$3x+3y-z+l=0$$, so

$$3\alpha+3\beta-2+l=0.$$

Using $$\beta=-\alpha-1$$:
$$3\alpha+3(-\alpha-1)-2+l=0 \;\Longrightarrow\;-5+l=0 \;\Longrightarrow\;l=5.$$

Therefore, the required value is 5.

We are given $$A(1, -1, 2)$$, $$B(5, 7, -6)$$, $$C(3, 4, -10)$$, $$D(-1, -4, -2)$$.

First, let us check whether $$ABCD$$ is a parallelogram by computing the midpoints of the diagonals.

Midpoint of $$AC = \left(\frac{1+3}{2}, \frac{-1+4}{2}, \frac{2+(-10)}{2}\right) = (2, 1.5, -4)$$

Midpoint of $$BD = \left(\frac{5+(-1)}{2}, \frac{7+(-4)}{2}, \frac{-6+(-2)}{2}\right) = (2, 1.5, -4)$$

Since both diagonals have the same midpoint, $$ABCD$$ is a parallelogram.

For a parallelogram, the area can be computed using the diagonal vectors with the formula:

$$\text{Area} = \frac{1}{2}|\overrightarrow{AC} \times \overrightarrow{BD}|$$

Computing the diagonal vectors:

$$\overrightarrow{AC} = C - A = (3-1, 4-(-1), -10-2) = (2, 5, -12)$$

$$\overrightarrow{BD} = D - B = (-1-5, -4-7, -2-(-6)) = (-6, -11, 4)$$

Now compute the cross product $$\overrightarrow{AC} \times \overrightarrow{BD}$$:

$$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & -12 \\ -6 & -11 & 4 \end{vmatrix}$$

$$\hat{i}$$-component: $$5 \times 4 - (-12) \times (-11) = 20 - 132 = -112$$

$$\hat{j}$$-component: $$-(2 \times 4 - (-12) \times (-6)) = -(8 - 72) = 64$$

$$\hat{k}$$-component: $$2 \times (-11) - 5 \times (-6) = -22 + 30 = 8$$

So $$\overrightarrow{AC} \times \overrightarrow{BD} = (-112, 64, 8)$$.

$$|\overrightarrow{AC} \times \overrightarrow{BD}| = \sqrt{(-112)^2 + 64^2 + 8^2}$$

$$= \sqrt{12544 + 4096 + 64}$$

$$= \sqrt{16704}$$

Let us simplify $$\sqrt{16704}$$. We find: $$16704 = 576 \times 29$$, since $$576 = 24^2$$ and $$576 \times 29 = 16704$$.

$$\sqrt{16704} = 24\sqrt{29}$$

Therefore, the area of the parallelogram $$ABCD$$ is:

$$\text{Area} = \frac{1}{2} \times 24\sqrt{29} = 12\sqrt{29}$$

The answer is Option B: $$12\sqrt{29}$$.

We are given $$A(3, 1, -1)$$, $$B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right)$$, $$C(2, 2, 1)$$, $$D\left(\frac{10}{3}, \frac{2}{3}, -\frac{1}{3}\right)$$.

First, let us check whether $$ABCD$$ is a parallelogram by computing the side vectors:

$$\overrightarrow{AB} = B - A = \left(\frac{5}{3} - 3, \frac{7}{3} - 1, \frac{1}{3} + 1\right) = \left(-\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right)$$

$$\overrightarrow{DC} = C - D = \left(2 - \frac{10}{3}, 2 - \frac{2}{3}, 1 + \frac{1}{3}\right) = \left(-\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right)$$

Since $$\overrightarrow{AB} = \overrightarrow{DC}$$, the quadrilateral $$ABCD$$ is a parallelogram.

For a parallelogram with adjacent sides $$\overrightarrow{AB}$$ and $$\overrightarrow{AD}$$, the area is $$|\overrightarrow{AB} \times \overrightarrow{AD}|$$.

Computing $$\overrightarrow{AD}$$:

$$\overrightarrow{AD} = D - A = \left(\frac{10}{3} - 3, \frac{2}{3} - 1, -\frac{1}{3} + 1\right) = \left(\frac{1}{3}, -\frac{1}{3}, \frac{2}{3}\right)$$

Computing the cross product $$\overrightarrow{AB} \times \overrightarrow{AD}$$:

$$\overrightarrow{AB} \times \overrightarrow{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -\frac{4}{3} & \frac{4}{3} & \frac{4}{3} \\ \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \end{vmatrix}$$

$$\hat{i}$$: $$\frac{4}{3} \cdot \frac{2}{3} - \frac{4}{3} \cdot \left(-\frac{1}{3}\right) = \frac{8}{9} + \frac{4}{9} = \frac{12}{9} = \frac{4}{3}$$

$$\hat{j}$$: $$-\left[\left(-\frac{4}{3}\right)\left(\frac{2}{3}\right) - \frac{4}{3} \cdot \frac{1}{3}\right] = -\left[-\frac{8}{9} - \frac{4}{9}\right] = -\left(-\frac{12}{9}\right) = \frac{4}{3}$$

$$\hat{k}$$: $$\left(-\frac{4}{3}\right)\left(-\frac{1}{3}\right) - \frac{4}{3} \cdot \frac{1}{3} = \frac{4}{9} - \frac{4}{9} = 0$$

So $$\overrightarrow{AB} \times \overrightarrow{AD} = \left(\frac{4}{3}, \frac{4}{3}, 0\right)$$.

$$|\overrightarrow{AB} \times \overrightarrow{AD}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + 0} = \frac{4}{3}\sqrt{2}$$

Therefore, the area of the parallelogram $$ABCD$$ is:

$$\text{Area} = \frac{4\sqrt{2}}{3}$$

The answer is Option D: $$\frac{4\sqrt{2}}{3}$$.

Statement I: $$\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$$, $$\vec{b}=2\hat{i}+\hat{j}-\hat{k}$$. $$\vec{a}\times\vec{r}=\vec{a}\times\vec{b}$$ means $$\vec{a}\times(\vec{r}-\vec{b})=\vec{0}$$, so $$\vec{r}-\vec{b}=\lambda\vec{a}$$.

$$\vec{r}=\vec{b}+\lambda\vec{a}=(2+\lambda)\hat{i}+(1+2\lambda)\hat{j}+(-1-3\lambda)\hat{k}$$.

$$\vec{a}\cdot\vec{r}=0$$: $$(2+\lambda)+2(1+2\lambda)-3(-1-3\lambda)=0$$. $$2+\lambda+2+4\lambda+3+9\lambda=0$$. $$14\lambda+7=0$$, $$\lambda=-1/2$$.

$$\vec{r}=3/2\hat{i}+0\hat{j}+1/2\hat{k}$$. $$|\vec{r}|=\sqrt{9/4+1/4}=\sqrt{10/4}=\sqrt{10}/2$$.

$$|\vec{r}| = \sqrt{10}/2 \neq \sqrt{10}$$. Statement I is incorrect.

Statement II: $$\cos 2A+\cos 2B+\cos 2C \geq -3/2$$.

$$\cos 2A+\cos 2B+\cos 2C = -1-4\cos A\cos B\cos C$$. Min of $$-1-4\cos A\cos B\cos C$$: max of $$\cos A\cos B\cos C = 1/8$$ (at equilateral). Min = $$-1-4(1/8) = -3/2$$.

So $$\geq -3/2$$. Statement II is correct.

The correct answer is Option 1.

Given $$|\vec{a}| = 2$$, $$|\vec{b}| = 3$$, $$\vec{a} = \vec{b} \times \vec{c}$$, and $$\alpha \in [0, \pi/3]$$ is the angle between $$\vec{b}$$ and $$\vec{c}$$.

Since $$\vec{a} = \vec{b} \times \vec{c}$$, we have $$|\vec{a}| = |\vec{b}||\vec{c}|\sin\alpha$$.

Also, $$\vec{a} \perp \vec{b}$$ and $$\vec{a} \perp \vec{c}$$.

Compute $$|\vec{c} - \vec{a}|^2$$.

Since $$\vec{a} \perp \vec{c}$$: $$\vec{a} \cdot \vec{c} = 0$$.

Minimize over $$\alpha \in [0, \pi/3]$$.

$$\frac{12}{\sin^2\alpha}$$ is minimized when $$\sin^2\alpha$$ is maximized. On $$[0, \pi/3]$$, $$\sin\alpha$$ is increasing, so maximum at $$\alpha = \pi/3$$: $$\sin(\pi/3) = \frac{\sqrt{3}}{2}$$, $$\sin^2(\pi/3) = \frac{3}{4}$$.

The correct answer is Option (2): 124.

$$\vec{AC} = (-3-2)\hat{i} + (4-3)\hat{j} + (-2-5)\hat{k} = -5\hat{i} + \hat{j} - 7\hat{k}$$

Area $$= \frac{1}{2} |\vec{AC} \times \vec{BD}|$$

$$\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -7 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3+14) - \hat{j}(-15+7) + \hat{k}(-10-1) = 17\hat{i} + 8\hat{j} - 11\hat{k}$$

Magnitude $$= \sqrt{17^2 + 8^2 + (-11)^2} = \sqrt{289 + 64 + 121} = \sqrt{474}$$

Area $$= \mathbf{\frac{1}{2}\sqrt{474}}$$ (Option B)

$$\vec{c} = \vec{a} - \vec{b} = (\alpha-5)\hat{i} + \hat{j} - 2\hat{k}$$.

Area of triangle = $$\frac{1}{2}|\vec{a} \times \vec{b}| = 5\sqrt{6}$$, so $$|\vec{a} \times \vec{b}| = 10\sqrt{6}$$.

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ 5 & 3 & 4 \end{vmatrix} = (16-6)\hat{i} - (4\alpha-10)\hat{j} + (3\alpha-20)\hat{k}$$

$$= 10\hat{i} - (4\alpha-10)\hat{j} + (3\alpha-20)\hat{k}$$

$$|\vec{a} \times \vec{b}|^2 = 100 + (4\alpha-10)^2 + (3\alpha-20)^2 = 600$$

$$(4\alpha-10)^2 + (3\alpha-20)^2 = 500$$

$$16\alpha^2 - 80\alpha + 100 + 9\alpha^2 - 120\alpha + 400 = 500$$

$$25\alpha^2 - 200\alpha + 500 = 500$$

$$25\alpha^2 - 200\alpha = 0 \implies 25\alpha(\alpha - 8) = 0$$

Since $$\alpha > 0$$: $$\alpha = 8$$.

$$\vec{c} = (8-5)\hat{i} + \hat{j} - 2\hat{k} = 3\hat{i} + \hat{j} - 2\hat{k}$$.

$$|\vec{c}|^2 = 9 + 1 + 4 = 14$$.

The correct answer is Option 2: 14.

Given $$\vec{a}=2\,\hat{i}+\,\hat{j}-\,\hat{k}$$ and $$\vec{b}=((\vec{a}\times(\hat{i}+\hat{j}))\times\hat{i})\times\hat{i}$$. We first evaluate the cross products one by one.

Step 1: Compute $$\vec{c}=\vec{a}\times(\hat{i}+\hat{j})$$.
Write $$\hat{i}+\hat{j}=(1,1,0)$$ and $$\vec{a}=(2,1,-1)$$.
Using the determinant formula for $$\vec{u}\times\vec{v}$$ :

$$\vec{c}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & -1\\ 1 & 1 & 0 \end{vmatrix} =\hat{i}(1\cdot0-(-1)\cdot1)-\hat{j}(2\cdot0-(-1)\cdot1)+\hat{k}(2\cdot1-1\cdot1)$$

$$\Rightarrow\;\vec{c}=1\,\hat{i}-1\,\hat{j}+1\,\hat{k}=(1,-1,1)$$.

Step 2: Compute $$\vec{d}=\vec{c}\times\hat{i}$$.
Take $$\hat{i}=(1,0,0)$$:

$$\vec{d}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -1 & 1\\ 1 & 0 & 0 \end{vmatrix} =\hat{i}((-1)\cdot0-1\cdot0)-\hat{j}(1\cdot0-1\cdot1)+\hat{k}(1\cdot0-(-1)\cdot1)$$

$$\Rightarrow\;\vec{d}=0\,\hat{i}+1\,\hat{j}+1\,\hat{k}=(0,1,1)$$.

Step 3: Compute $$\vec{b}=\vec{d}\times\hat{i}$$.
Again cross with $$\hat{i}=(1,0,0)$$ :

$$\vec{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 1 & 1\\ 1 & 0 & 0 \end{vmatrix} =\hat{i}(1\cdot0-1\cdot0)-\hat{j}(0\cdot0-1\cdot1)+\hat{k}(0\cdot0-1\cdot1)$$

$$\Rightarrow\;\vec{b}=0\,\hat{i}+1\,\hat{j}-1\,\hat{k}=(0,1,-1)$$.

Step 4: Square of the projection of $$\vec{a}$$ on $$\vec{b}$$.
Projection length formula: the component of $$\vec{a}$$ along $$\vec{b}$$ is $$\dfrac{\vec{a}\cdot\vec{b}}{\lvert\vec{b}\rvert}$$. Hence

$$\text{(projection)}^{2}=\dfrac{(\vec{a}\cdot\vec{b})^{2}}{\lvert\vec{b}\rvert^{2}}$$.

Dot product: $$\vec{a}\cdot\vec{b}=(2,1,-1)\cdot(0,1,-1)=0+1+1=2$$.
Magnitude squared of $$\vec{b}$$: $$\lvert\vec{b}\rvert^{2}=0^{2}+1^{2}+(-1)^{2}=2$$.

Therefore $$\text{(projection)}^{2}=\dfrac{2^{2}}{2}=2$$.

Hence the square of the projection of $$\vec{a}$$ on $$\vec{b}$$ is $$2$$, which matches Option C.

$$\vec{p}\times\vec{b}=\vec{c}\times\vec{b}$$, so $$(\vec{p}-\vec{c})\times\vec{b}=0$$, meaning $$\vec{p}-\vec{c}=\lambda\vec{b}$$.

$$\vec{p}=\vec{c}+\lambda\vec{b}=(1+4\lambda)\hat{i}+(-3+\lambda)\hat{j}+(4+7\lambda)\hat{k}$$.

$$\vec{p}\cdot\vec{a}=0$$: $$3(1+4\lambda)+1(-3+\lambda)-2(4+7\lambda)=0$$.

$$3+12\lambda-3+\lambda-8-14\lambda=0 \Rightarrow -\lambda-8=0 \Rightarrow \lambda=-8$$.

$$\vec{p}=(-31)\hat{i}+(-11)\hat{j}+(-52)\hat{k}$$.

$$\vec{p}\cdot(\hat{i}-\hat{j}-\hat{k})=-31+11+52=32$$.

Given $$\vec{a} = 4\hat{i} - \hat{j} + \hat{k}$$, $$\vec{b} = 11\hat{i} - \hat{j} + \hat{k}$$, and $$(\vec{a}+\vec{b}) \times \vec{c} = \vec{c} \times(-2\vec{a}+3\vec{b})$$.

We use the fact that

$$\vec{c} \times(-2\vec{a}+3\vec{b}) = -(-2\vec{a}+3\vec{b}) \times \vec{c}$$.

Hence

$$ (\vec{a}+\vec{b}) \times \vec{c} + (-2\vec{a}+3\vec{b}) \times \vec{c} = \vec{0} $$

Applying the distributive property of the cross product leads to

$$ ((\vec{a}+\vec{b}) + (-2\vec{a}+3\vec{b})) \times \vec{c} = \vec{0} $$

which simplifies to

$$ (-\vec{a}+4\vec{b}) \times \vec{c} = \vec{0}. $$

Therefore, $$\vec{c}$$ must be parallel to $$(-\vec{a}+4\vec{b})$$.

Substituting the component forms gives

$$ -\vec{a} + 4\vec{b} = -(4\hat{i}-\hat{j}+\hat{k}) + 4(11\hat{i}-\hat{j}+\hat{k}) = -4\hat{i}+\hat{j}-\hat{k}+44\hat{i}-4\hat{j}+4\hat{k} = 40\hat{i}-3\hat{j}+3\hat{k}. $$

Thus we can write $$\vec{c} = t(40\hat{i}-3\hat{j}+3\hat{k})$$ for some scalar $$t$$.

Using the condition $$(2\vec{a}+3\vec{b}) \cdot \vec{c} = 1670$$ we compute

$$ 2\vec{a} + 3\vec{b} = 2(4\hat{i}-\hat{j}+\hat{k}) + 3(11\hat{i}-\hat{j}+\hat{k}) = 8\hat{i}-2\hat{j}+2\hat{k}+33\hat{i}-3\hat{j}+3\hat{k} = 41\hat{i}-5\hat{j}+5\hat{k}. $$

Substitution into the dot product yields

$$ (41\hat{i}-5\hat{j}+5\hat{k}) \cdot t(40\hat{i}-3\hat{j}+3\hat{k}) = 1670. $$

This simplifies to

$$ t(41 \times 40 + (-5)(-3) + 5 \times 3) = 1670, $$

so

$$ t(1640 + 15 + 15) = 1670, $$

giving

$$ t \times 1670 = 1670 \implies t = 1. $$

Therefore $$\vec{c} = 40\hat{i}-3\hat{j}+3\hat{k}$$ and its squared magnitude is

$$ |\vec{c}|^2 = 40^2 + 3^2 + 3^2 = 1600 + 9 + 9 = 1618. $$

The correct answer is Option 2: 1618.

$$|\vec{a}|^2 = 1 + \alpha^2 + \beta^2$$. $$|\vec{b}|^2 = 6$$.

$$\vec{a} \cdot \vec{b} = 3\sqrt{2}$$.

Angle $$\pi/4$$ between $$\vec{a}$$ and $$\vec{b}$$: $$\cos(\pi/4) = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{3\sqrt{2}}{|\vec{a}|\sqrt{6}}$$.

$$\frac{1}{\sqrt{2}} = \frac{3\sqrt{2}}{|\vec{a}|\sqrt{6}} \Rightarrow |\vec{a}|\sqrt{6} = 3\sqrt{2} \cdot \sqrt{2} = 6 \Rightarrow |\vec{a}| = \sqrt{6}$$.

So $$1 + \alpha^2 + \beta^2 = 6 \Rightarrow \alpha^2 + \beta^2 = 5$$.

$$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 36 - 18 = 18$$.

$$(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2 = 5 \times 18 = 90$$.

$$\vec{a} + 5\vec{b}$$ is collinear with $$\vec{c}$$, and $$\vec{b} + 6\vec{c}$$ is collinear with $$\vec{a}$$. Also $$\vec{a} + \alpha\vec{b} + \beta\vec{c} = \vec{0}$$.

Express the collinearity conditions.

Since $$\vec{a} + 5\vec{b}$$ is collinear with $$\vec{c}$$:

$$\vec{a} + 5\vec{b} = \lambda\vec{c}$$ for some scalar $$\lambda$$ ... (i)

Since $$\vec{b} + 6\vec{c}$$ is collinear with $$\vec{a}$$:

$$\vec{b} + 6\vec{c} = \mu\vec{a}$$ for some scalar $$\mu$$ ... (ii)

Express $$\vec{a}$$ from equation (ii).

$$\vec{a} = \frac{1}{\mu}\vec{b} + \frac{6}{\mu}\vec{c}$$

Substitute into equation (i).

$$\frac{1}{\mu}\vec{b} + \frac{6}{\mu}\vec{c} + 5\vec{b} = \lambda\vec{c}$$

$$\left(\frac{1}{\mu} + 5\right)\vec{b} + \left(\frac{6}{\mu} - \lambda\right)\vec{c} = \vec{0}$$

Since $$\vec{b}$$ and $$\vec{c}$$ are non-collinear, both coefficients must be zero:

$$\frac{1}{\mu} + 5 = 0 \implies \mu = -\frac{1}{5}$$

$$\frac{6}{\mu} = \lambda \implies \lambda = 6 \times (-5) = -30$$

Find $$\alpha$$ and $$\beta$$.

From (i): $$\vec{a} = -5\vec{b} + \lambda\vec{c} = -5\vec{b} - 30\vec{c}$$

So $$\vec{a} + 5\vec{b} + 30\vec{c} = \vec{0}$$

Therefore $$\alpha = 5$$ and $$\beta = 30$$.

$$\alpha + \beta = 5 + 30 = 35$$

The answer is Option 1: 35.

$$\vec{OA} = \vec{a}$$, $$\vec{OB} = 12\vec{a} + 4\vec{b}$$, $$\vec{OC} = \vec{b}$$, where $$O$$ is the origin.

$$S$$ has adjacent sides $$\vec{OA} = \vec{a}$$ and $$\vec{OC} = \vec{b}$$:

$$\text{Area of } S = |\vec{a} \times \vec{b}|$$

The area of quadrilateral $$OABC$$ with vertices in order $$O, A, B, C$$ can be computed by splitting into triangles $$OAB$$ and $$OBC$$:

$$\text{Area}(OABC) = \text{Area}(OAB) + \text{Area}(OBC)$$

$$\text{Area}(OAB) = \frac{1}{2}|\vec{OA} \times \vec{OB}| = \frac{1}{2}|\vec{a} \times (12\vec{a} + 4\vec{b})| = \frac{1}{2}|12(\vec{a} \times \vec{a}) + 4(\vec{a} \times \vec{b})| =  2|\vec{a} \times \vec{b}|$$

$$\text{Area}(OBC) = \frac{1}{2}|\vec{OB} \times \vec{OC}| = \frac{1}{2}|(12\vec{a} + 4\vec{b}) \times \vec{b}| = \frac{1}{2}|12(\vec{a} \times \vec{b}) + 4(\vec{b} \times \vec{b})| = 6|\vec{a} \times \vec{b}|$$

$$\text{Area}(OABC) = 2|\vec{a} \times \vec{b}| + 6|\vec{a} \times \vec{b}| = 8|\vec{a} \times \vec{b}|$$

$$\frac{\text{Area}(OABC)}{\text{Area}(S)} = \frac{8|\vec{a} \times \vec{b}|}{|\vec{a} \times \vec{b}|} = 8$$

We are given
$$\vec{a}=1\,\hat{i}+2\,\hat{j}+1\,\hat{k},\qquad \vec{b}=3(\hat{i}-\hat{j}+\hat{k})=3\,\hat{i}-3\,\hat{j}+3\,\hat{k}.$$

Let $$\vec{c}=x\,\hat{i}+y\,\hat{j}+z\,\hat{k}.$$
The two conditions on $$\vec{c}$$ are

1. $$\vec{a}\times\vec{c}=\vec{b}$$

2. $$\vec{a}\cdot\vec{c}=3$$

Case 1: Solve $$\vec{a}\times\vec{c}=\vec{b}.$$ Using the determinant rule for a cross product,

$$\vec{a}\times\vec{c}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 1&2&1\\ x&y&z \end{vmatrix} =\hat{i}(2z-y)-\hat{j}(z-x)+\hat{k}(y-2x).$$

Equating components with $$\vec{b}=(3,\,-3,\,3),$$ we obtain

$$2z-y=3\quad -(1)$$

$$-(z-x)=-3\;\Longrightarrow\; z-x=3\quad -(2)$$

$$y-2x=3\quad -(3)$$

Case 2: Solve $$\vec{a}\cdot\vec{c}=3.$$ The dot product gives

$$1\cdot x+2\cdot y+1\cdot z=3\quad -(4)$$

From $$(2)$$ we get $$x=z-3.$$
From $$(1)$$ we get $$y=2z-3.$$

Substitute these in $$(4):$$

$$(z-3)+2(2z-3)+z=3$$

$$z-3+4z-6+z=3$$

$$6z-9=3$$

$$6z=12\;\Longrightarrow\; z=2.$$

Hence

$$x=z-3=2-3=-1,\qquad y=2z-3=4-3=1.$$

Therefore $$\vec{c}=-1\,\hat{i}+1\,\hat{j}+2\,\hat{k}.$$

Case 3: Evaluate the required expression $$\vec{a}\cdot\bigl((\vec{c}\times\vec{b})-\vec{b}-\vec{c}\bigr).$$

First compute $$\vec{c}\times\vec{b}:$$

$$\vec{c}\times\vec{b}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ -1&1&2\\ 3&-3&3 \end{vmatrix} =\hat{i}(1\cdot3-2\cdot(-3)) -\hat{j}((-1)\cdot3-2\cdot3) +\hat{k}((-1)(-3)-1\cdot3)$$

$$=\hat{i}(3+6)-\hat{j}(-3-6)+\hat{k}(3-3) =9\,\hat{i}+9\,\hat{j}+0\,\hat{k}.$$

Now form the vector inside the parentheses:

$$(\vec{c}\times\vec{b})-\vec{b}-\vec{c} =(9,9,0)-(3,-3,3)-(-1,1,2)$$

$$=(9-3+1,\;9+3-1,\;0-3-2) =(7,\;11,\;-5).$$

Finally take the dot product with $$\vec{a}=(1,2,1)\!:\!$$

$$\vec{a}\cdot(7,11,-5)=1\cdot7+2\cdot11+1\cdot(-5)=7+22-5=24.$$

Hence the value of $$\vec{a}\cdot\bigl((\vec{c}\times\vec{b})-\vec{b}-\vec{c}\bigr)=24.$$

Therefore, the correct option is Option B (24).

For the angle $$\frac{\pi}{2}$$ with $$\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$$, we have $$ \frac{x + z}{\sqrt{2}} = \cos\frac{\pi}{2} = 0 \implies x + z = 0 \quad \cdots(1) $$.

For the angle $$\frac{\pi}{3}$$ with $$\frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$$, $$ \frac{y + z}{\sqrt{2}} = \cos\frac{\pi}{3} = \frac{1}{2} \implies y + z = \frac{1}{\sqrt{2}} \quad \cdots(2)$$.

For the angle $$\frac{2\pi}{3}$$ with $$\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$$, $$ \frac{x + y}{\sqrt{2}} = \cos\frac{2\pi}{3} = -\frac{1}{2} \implies x + y = -\frac{1}{\sqrt{2}} \quad \cdots(3)$$.

From (1), $$z = -x$$. Substituting into (2) gives $$y - x = \frac{1}{\sqrt{2}} \quad \cdots(4)$$, and (3) states $$x + y = -\frac{1}{\sqrt{2}} \quad \cdots(5)$$. 

Adding (4) and (5) yields $$2y = 0 \implies y = 0$$. Then (5) implies $$x = -\frac{1}{\sqrt{2}}$$ and (1) gives $$z = \frac{1}{\sqrt{2}}$$, so $$\hat{u} = -\frac{1}{\sqrt{2}}\hat{i} + 0\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$$.

To compute $$|\hat{u} - \vec{v}|^2$$ where $$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$$, note that

$$\hat{u} - \vec{v} = \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{i} + \left(0 - \frac{1}{\sqrt{2}}\right)\hat{j} + \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{k}$$

$$= -\sqrt{2}\,\hat{i} - \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k}$$

$$ |\hat{u} - \vec{v}|^2 = 2 + \frac{1}{2} + 0 = \frac{5}{2} $$. Thus the correct answer is Option (2): $$\frac{5}{2}$$.

We need to find a unit vector $$\overrightarrow{C} = x\hat{i} + y\hat{j} + z\hat{k}$$ such that it makes an angle of 60° with $$\overrightarrow{a} = 2\hat{i} + 2\hat{j} - \hat{k}$$ and an angle of 45° with $$\overrightarrow{b} = \hat{i} - \hat{k}$$.

Since $$\overrightarrow{C}$$ is a unit vector, we have:

$$ x^2 + y^2 + z^2 = 1 \quad \cdots(1) $$

Condition 1: Angle with $$\overrightarrow{a}$$ is 60°.
$$|\overrightarrow{a}| = \sqrt{4 + 4 + 1} = 3$$
$$ \cos 60° = \frac{\overrightarrow{C} \cdot \overrightarrow{a}}{|\overrightarrow{C}||\overrightarrow{a}|} = \frac{2x + 2y - z}{3} $$
$$ \frac{1}{2} = \frac{2x + 2y - z}{3} \implies 2x + 2y - z = \frac{3}{2} \quad \cdots(2) $$

Condition 2: Angle with $$\overrightarrow{b}$$ is 45°.
$$|\overrightarrow{b}| = \sqrt{1 + 1} = \sqrt{2}$$
$$ \cos 45° = \frac{\overrightarrow{C} \cdot \overrightarrow{b}}{|\overrightarrow{C}||\overrightarrow{b}|} = \frac{x - z}{\sqrt{2}} $$
$$ \frac{1}{\sqrt{2}} = \frac{x - z}{\sqrt{2}} \implies x - z = 1 \quad \cdots(3) $$

From (3): $$z = x - 1$$. Substituting into (2):
$$ 2x + 2y - (x - 1) = \frac{3}{2} \implies x + 2y + 1 = \frac{3}{2} \implies x = \frac{1}{2} - 2y \quad \cdots(4) $$

Also, $$z = x - 1 = -\frac{1}{2} - 2y$$. Substituting into (1):

$$ \left(\frac{1}{2} - 2y\right)^2 + y^2 + \left(-\frac{1}{2} - 2y\right)^2 = 1 $$

$$ \frac{1}{4} - 2y + 4y^2 + y^2 + \frac{1}{4} + 2y + 4y^2 = 1 $$

$$ 9y^2 + \frac{1}{2} = 1 \implies y^2 = \frac{1}{18} \implies y = \pm \frac{1}{3\sqrt{2}} $$

Case 1: $$y = \frac{1}{3\sqrt{2}}$$, then $$x = \frac{1}{2} - \frac{\sqrt{2}}{3}$$, $$z = -\frac{1}{2} - \frac{\sqrt{2}}{3}$$.
Case 2: $$y = -\frac{1}{3\sqrt{2}}$$, then $$x = \frac{1}{2} + \frac{\sqrt{2}}{3}$$, $$z = -\frac{1}{2} + \frac{\sqrt{2}}{3}$$.

Now compute $$\overrightarrow{C} + \left(-\frac{1}{2}\hat{i} + \frac{1}{3\sqrt{2}}\hat{j} - \frac{\sqrt{2}}{3}\hat{k}\right)$$ for each case.
Case 2 gives $$\frac{\sqrt{2}}{3}\hat{i} - \frac{1}{2}\hat{k}$$, which matches Option 1..

$$\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$$, so $$|\vec{OA}| = \sqrt{4+4+1} = 3$$

$$\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$$, so $$|\vec{OB}| = \sqrt{4+16+16} = 6$$

Apply the angle bisector theorem.

By the angle bisector theorem, the internal bisector of $$\angle AOB$$ divides $$AB$$ in the ratio $$OA:OB = 3:6 = 1:2$$.

Using the section formula (internal division in ratio $$1:2$$):

$$\vec{OC} = \frac{1 \cdot \vec{OB} + 2 \cdot \vec{OA}}{1 + 2} = \frac{(2\hat{i}+4\hat{j}+4\hat{k}) + 2(2\hat{i}+2\hat{j}+\hat{k})}{3}$$

$$= \frac{(2+4)\hat{i} + (4+4)\hat{j} + (4+2)\hat{k}}{3} = \frac{6\hat{i} + 8\hat{j} + 6\hat{k}}{3} = 2\hat{i} + \frac{8}{3}\hat{j} + 2\hat{k}$$

$$|OC| = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72 + 64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{136}}{3} = \frac{2\sqrt{34}}{3}$$

$$A(2,2,1)$$, $$B(1,2,2)$$, $$C(2,1,2)$$.

$$AB = \sqrt{1+0+1} = \sqrt{2}$$, $$BC = \sqrt{1+1+0} = \sqrt{2}$$, $$CA = \sqrt{0+1+1} = \sqrt{2}$$.

The triangle is equilateral. For an equilateral triangle, the orthocenter coincides with the centroid.

Centroid/orthocenter: $$H = \left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$$.

For an equilateral triangle with side $$a = \sqrt{2}$$, the distance from the center to any side is $$h/3$$ where $$h = \frac{\sqrt{3}}{2}a$$ is the height. So distance = $$\frac{a}{2\sqrt{3}} = \frac{\sqrt{2}}{2\sqrt{3}} = \frac{1}{\sqrt{6}}$$.

All three perpendicular distances are equal: $$l_1 = l_2 = l_3 = \frac{1}{\sqrt{6}}$$.

$$l_1^2 + l_2^2 + l_3^2 = 3 \times \frac{1}{6} = \frac{1}{2}$$.

The answer corresponds to Option (2).

Given $$\vec{a} = 2\hat{i} + 5\hat{j} - \hat{k}$$, $$\vec{b} = 2\hat{i} - 2\hat{j} + 2\hat{k}$$, and:

Let $$\vec{d} = \vec{c} + \hat{i}$$. Using $$\vec{a} \times \vec{d} = -\vec{d} \times \vec{a}$$:

So $$\vec{d}$$ is parallel to $$2\vec{a} + \vec{b} + \hat{i}$$.

$$2\vec{a} + \vec{b} + \hat{i} = 7\hat{i} + 8\hat{j}$$.

So $$\vec{d} = \lambda(7\hat{i} + 8\hat{j})$$ and $$\vec{c} = (7\lambda - 1)\hat{i} + 8\lambda\hat{j}$$.

Using $$\vec{a} \cdot \vec{c} = -29$$:

$$\vec{c} = -\frac{9}{2}\hat{i} - 4\hat{j}$$.

$$\vec{c} \cdot (-2\hat{i} + \hat{j} + \hat{k}) = 9 - 4 + 0 = 5$$.

The correct answer is Option 4: 5.

Given: $$\vec{a} = 2\hat{i} + \alpha\hat{j} + \hat{k}$$, $$\vec{b} = -\hat{i} + \hat{k}$$, $$\vec{c} = \beta\hat{j} - \hat{k}$$, with $$\alpha\beta = -6$$.

Diagonals: $$\vec{a} + \vec{b} = (1, \alpha, 2)$$ and $$\vec{b} + \vec{c} = (-1, \beta, 0)$$.

Area of parallelogram with diagonals $$\vec{d_1}$$ and $$\vec{d_2}$$ = $$\frac{1}{2}|\vec{d_1} \times \vec{d_2}|$$.

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 2 \\ -1 & \beta & 0 \end{vmatrix} = \hat{i}(0-2\beta) - \hat{j}(0+2) + \hat{k}(\beta+\alpha)$$

$$= (-2\beta, -2, \alpha+\beta)$$

$$|\vec{d_1} \times \vec{d_2}|^2 = 4\beta^2 + 4 + (\alpha+\beta)^2$$

Area = $$\frac{\sqrt{21}}{2}$$, so $$\frac{1}{4}[4\beta^2 + 4 + (\alpha+\beta)^2] = \frac{21}{4}$$.

$$4\beta^2 + 4 + \alpha^2 + 2\alpha\beta + \beta^2 = 21$$

$$\alpha^2 + 5\beta^2 + 2(-6) + 4 = 21$$ (using $$\alpha\beta = -6$$)

$$\alpha^2 + 5\beta^2 = 29$$

With $$\alpha\beta = -6$$ (integer solutions): possible $$(\alpha, \beta)$$ pairs: $$(1$$, $$-6)$$, $$(-1$$, $$6)$$, $$(2$$, $$-3)$$, $$(-2$$, $$3)$$, $$(3$$, $$-2)$$, $$(-3$$, $$2)$$, $$(6$$, $$-1)$$, $$(-6$$, $$1)$$.

Check $$\alpha^2 + 5\beta^2 = 29$$:

$$(\alpha_1, \beta_1) = (3, -2)$$ and $$(\alpha_2, \beta_2) = (-3, 2)$$.

$$\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2 = 9 + 4 - (-3)(2) = 13 + 6 = 19$$.

The correct answer is Option A: 19

Given vectors are:

$$\vec{a} = -5\hat{i} + \hat{j} - 3\hat{k}$$

$$\vec{b} = \hat{i} + 2\hat{j} - 4\hat{k}$$

We need to find $$\vec{c} = ((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i}$$ and then compute $$\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$$.

First, compute $$\vec{d} = \vec{a} \times \vec{b}$$ using the cross product formula:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -3 \\ 1 & 2 & -4 \\ \end{vmatrix}$$

Expanding the determinant:

$$\hat{i} \left( (1)(-4) - (-3)(2) \right) - \hat{j} \left( (-5)(-4) - (-3)(1) \right) + \hat{k} \left( (-5)(2) - (1)(1) \right)$$

Calculate each component:

• $$\hat{i}$$ component: $$(1)(-4) - (-3)(2) = -4 + 6 = 2$$
• $$\hat{j}$$ component: $$- \left[ (-5)(-4) - (-3)(1) \right] = - \left[ 20 + 3 \right] = -23$$
• $$\hat{k}$$ component: $$(-5)(2) - (1)(1) = -10 - 1 = -11$$

So, $$\vec{d} = 2\hat{i} - 23\hat{j} - 11\hat{k}$$

Next, compute $$\vec{e} = \vec{d} \times \hat{i}$$. The cross product of any vector $$\vec{u} = u_x \hat{i} + u_y \hat{j} + u_z \hat{k}$$ with $$\hat{i}$$ is given by:

$$\vec{u} \times \hat{i} = (u_z) \hat{j} + (-u_y) \hat{k}$$

Here, $$\vec{d} = 2\hat{i} - 23\hat{j} - 11\hat{k}$$, so $$u_x = 2$$, $$u_y = -23$$, $$u_z = -11$$.

Thus, $$\vec{e} = (-11) \hat{j} + (-(-23)) \hat{k} = -11\hat{j} + 23\hat{k}$$

Now, compute $$\vec{f} = \vec{e} \times \hat{i}$$ using the same formula:

$$\vec{e} = 0\hat{i} -11\hat{j} + 23\hat{k}$$, so $$u_x = 0$$, $$u_y = -11$$, $$u_z = 23$$.

Thus, $$\vec{f} = (23) \hat{j} + (-(-11)) \hat{k} = 23\hat{j} + 11\hat{k}$$

Finally, compute $$\vec{c} = \vec{f} \times \hat{i}$$:

$$\vec{f} = 0\hat{i} + 23\hat{j} + 11\hat{k}$$, so $$u_x = 0$$, $$u_y = 23$$, $$u_z = 11$$.

Thus, $$\vec{c} = (11) \hat{j} + (-23) \hat{k} = 11\hat{j} - 23\hat{k}$$

Now, compute the dot product $$\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$$:

$$\vec{c} = 0\hat{i} + 11\hat{j} - 23\hat{k}$$

Dot product: $$(0)(-1) + (11)(1) + (-23)(1) = 0 + 11 - 23 = -12$$

Therefore, the value is $$-12$$, which corresponds to option A.

|a| = √(36+1+1) = √38. a·c = 6|c|: cos θ_ac = 6/√38.

|c-a|² = |c|²-2a·c+|a|² = |c|²-12|c|+38 = 8. So |c|²-12|c|+30 = 0. |c| = (12±√(144-120))/2 = (12±√24)/2 = 6±√6. Since |c|≥6: |c| = 6+√6.

a×b = |i j k; 6 1 -1; 1 1 0| = (0+1)i-(-1+0... = i-(-1)j+(6-1)k... Let me compute: = (0+1)i-(0+1)j+(6-1)k = i-j+5k. |a×b| = √(1+1+25) = √27 = 3√3.

|(a×b)×c| = |a×b|·|c|·sin60° = 3√3·(6+√6)·(√3/2) = (9/2)(6+√6).

The correct answer is Option (3): 9(6+√6)/2.

Given $$|\vec{a}| = 1$$, $$\vec{a} \cdot \vec{b} = 2$$, $$|\vec{b}| = 4$$, and $$\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$$.

$$\vec{b} \cdot \vec{c} = 2\vec{b} \cdot (\vec{a} \times \vec{b}) - 3|\vec{b}|^2 = 0 - 3(16) = -48$$

(since $$\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$$)

$$|\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 - 12(\vec{a} \times \vec{b}) \cdot \vec{b} + 9|\vec{b}|^2$$

$$= 4|\vec{a} \times \vec{b}|^2 + 144$$

$$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = 16 - 4 = 12$$.

$$|\vec{c}|^2 = 48 + 144 = 192$$

$$\cos\theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|} = \frac{-48}{4\sqrt{192}} = \frac{-48}{4 \cdot 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = \frac{-\sqrt{3}}{2}$$

$$\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = 150°$$

The answer is Option (3): $$\boxed{\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)}$$.

$$|\vec{b}| = 1$$, $$|\vec{b} \times \vec{a}| = 2$$.

$$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 - 2(\vec{b} \times \vec{a}) \cdot \vec{b} + |\vec{b}|^2$$

Since $$(\vec{b} \times \vec{a}) \perp \vec{b}$$: $$(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$$.

$$= 4 + 0 + 1 = 5$$.

The answer is Option (2): $$\boxed{5}$$.

We are given $$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}, \quad \vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}$$.

We denote $$\vec{r}$$ as the unit vector along $$\vec{b} + \vec{c}$$ and use the condition $$\vec{r} \cdot \vec{a} = 3$$.

First, we compute $$\vec{b} + \vec{c}$$:

$$\vec{b}+\vec{c} = (2+3)\hat{i} + (3-1)\hat{j} + (-5+\lambda)\hat{k} = 5\hat{i} + 2\hat{j} + (\lambda-5)\hat{k}$$

Next, its magnitude is

$$|\vec{b} + \vec{c}| = \sqrt{25 + 4 + (\lambda - 5)^2} = \sqrt{29 + (\lambda - 5)^2}$$

Therefore the unit vector is

$$\vec{r} = \frac{5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}}{\sqrt{29 + (\lambda - 5)^2}}$$.

Using the dot product condition gives

$$\vec{r} \cdot \vec{a} = \frac{5(1) + 2(2) + (\lambda - 5)(3)}{\sqrt{29 + (\lambda - 5)^2}} = \frac{5 + 4 + 3\lambda - 15}{\sqrt{29 + (\lambda - 5)^2}} = \frac{3\lambda - 6}{\sqrt{29 + (\lambda - 5)^2}} = 3$$

This leads to the equation

$$3\lambda - 6 = 3\sqrt{29 + (\lambda - 5)^2}$$

Dividing both sides by 3 yields

$$\lambda - 2 = \sqrt{29 + (\lambda - 5)^2}$$

Since the square root is non-negative, we require $$\lambda \ge 2$$. Squaring both sides gives

$$ (\lambda - 2)^2 = 29 + (\lambda - 5)^2 $$

Expanding both sides results in

$$ \lambda^2 - 4\lambda + 4 = 29 + \lambda^2 - 10\lambda + 25 $$

which simplifies to

$$ -4\lambda + 4 = 54 - 10\lambda $$

and hence

$$ 6\lambda = 50 $$

so that

$$ \lambda = \frac{25}{3} $$

Therefore,

$$ 3\lambda = 25 $$.

The correct value of $$3\lambda$$ is 25.

Given $$\vec{a}=(1,1,1)$$, $$\vec{b}=(2,4,-5)$$, and $$\vec{c}=(x,2,3)$$, we seek $$(\vec{a}\times\vec{b})\cdot\vec{c}$$ under the constraint that $$\vec{d}$$ is a unit vector in the direction of $$\vec{b}+\vec{c}$$ and satisfies $$\vec{a}\cdot\vec{d}=1$$.

Since $$\vec{d} = \frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|}$$ and $$\vec{b}+\vec{c} = (2+x,6,-2)$$, we have $$\vec{a}\cdot\vec{d} = \frac{(1)(2+x)+(1)(6)+(1)(-2)}{|\vec{b}+\vec{c}|} = \frac{x+6}{|\vec{b}+\vec{c}|} = 1$$.

Since $$|\vec{b}+\vec{c}| = \sqrt{(2+x)^2 + 36 + 4} = \sqrt{(2+x)^2 + 40}$$, substituting into the previous equation gives $$x+6 = \sqrt{(x+2)^2 + 40}$$. Squaring both sides yields $$(x+6)^2 = (x+2)^2 + 40$$, which simplifies to $$x^2 + 12x + 36 = x^2 + 4x + 44$$. This leads to $$8x = 8$$ and hence $$x = 1$$.

With $$x = 1$$, the vector $$\vec{c}$$ becomes $$(1,2,3)$$. Therefore the scalar triple product is $$[\vec{a},\vec{b},\vec{c}] = \begin{vmatrix}1 & 1 & 1\\ 2 & 4 & -5\\ 1 & 2 & 3\end{vmatrix}$$. Expanding gives $$1(12+10) - 1(6+5) + 1(4-4) = 22 - 11 + 0 = 11$$.

The correct answer is Option (1): 11.

$$\vec{OA} = 2\vec{a}$$, $$\vec{OB} = 6\vec{a} + 5\vec{b}$$, $$\vec{OC} = 3\vec{b}$$.

Area of parallelogram with sides $$\vec{OA}$$ and $$\vec{OC}$$ = $$|2\vec{a} \times 3\vec{b}| = 6|\vec{a} \times \vec{b}| = 15$$.

So $$|\vec{a} \times \vec{b}| = \frac{15}{6} = \frac{5}{2}$$.

Area of quadrilateral OABC = Area of triangle OAB + Area of triangle OBC.

Area of $$\triangle OAB = \frac{1}{2}|\vec{OA} \times \vec{OB}| = \frac{1}{2}|2\vec{a} \times (6\vec{a}+5\vec{b})| = \frac{1}{2}|10(\vec{a}\times\vec{b})| = 5|\vec{a}\times\vec{b}| = \frac{25}{2}$$.

Area of $$\triangle OBC = \frac{1}{2}|\vec{OB} \times \vec{OC}| = \frac{1}{2}|(6\vec{a}+5\vec{b}) \times 3\vec{b}| = \frac{1}{2}|18(\vec{a}\times\vec{b})| = 9|\vec{a}\times\vec{b}| = \frac{45}{2}$$.

Total area = $$\frac{25}{2} + \frac{45}{2} = \frac{70}{2} = 35$$.

The correct answer is Option 4: 35.

Since $$(\vec{a} + \vec{b}) \perp (\vec{a} - \vec{b})$$, their dot product is zero: $$|\vec{a}|^2 = |\vec{b}|^2$$.

Equating magnitudes: $$1^2 + \lambda^2 + (-3)^2 = 3^2 + (-1)^2 + 2^2 \implies \lambda^2 + 10 = 14 \implies \lambda = 2$$ (since $$\lambda > 0$$).

Dot product $$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-1) + (-3)(2) = -5$$.

$$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-5}{\sqrt{14}\sqrt{14}} = -\frac{5}{14}$$.

Evaluating $$(14 \cos \theta)^2 = (-5)^2 = 25$$.

AB = i + 2j - 7k, AC = 6i + dj - 2k.

BC = AC - AB = 5i + (d-2)j + 5k.

Area = (1/2)|AB × AC| = 15√2, so |AB × AC| = 30√2.

AB × AC =  = (2(-2)-(-7)d)i - (1(-2)-(-7)(6))j + (1d-2(6))k

= (-4+7d)i - (-2+42)j + (d-12)k = (7d-4)i - 40j + (d-12)k

|AB × AC|² = (7d-4)² + 1600 + (d-12)² = 50d²-80d+1760= (30√2)² = 1800

50d²-80d+1760 = 1800 → 50d²-80d-40 = 0 → 5d²-8d-4 = 0

(5d+2)(d-2) = 0 → d = 2 (since d > 0)

|AB|² = 1+4+49 = 54, |BC|² = 25+0+25 = 50, |AC|² = 36+4+4 = 44

Largest side: AB² = 54

We are given
$$\vec a = 2\hat i - 3\hat j + 4\hat k,\; \vec b = 3\hat i + 4\hat j - 5\hat k,\; \vec c = x\hat i + y\hat j + z\hat k$$

The vector equation is
$$\vec a \times (\vec b + \vec c) + \vec b \times \vec c = \hat i + 8\hat j + 13\hat k \; -(1)$$

First expand the cross-product term:
$$\vec a \times (\vec b + \vec c)=\vec a \times \vec b+\vec a \times \vec c$$

Hence $$-(1)$$ becomes
$$\vec a \times \vec b + \vec a \times \vec c + \vec b \times \vec c = \hat i + 8\hat j + 13\hat k \; -(2)$$

Step 1 Compute $$\vec a \times \vec b$$ :

$$ \vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] 2 & -3 & 4\\[2pt] 3 & 4 & -5 \end{vmatrix} = \bigl((-3)(-5)-4\cdot4\bigr)\hat i - \bigl(2(-5)-4\cdot3\bigr)\hat j + \bigl(2\cdot4-(-3)\cdot3\bigr)\hat k$$ $$= (-1)\hat i + 22\hat j + 17\hat k$$ So $$\vec a \times \vec b = (-1,\,22,\,17).$$

Step 2 Write $$\vec a \times \vec c$$ :

$$ \vec a \times \vec c = \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] 2 & -3 & 4\\[2pt] x & y & z \end{vmatrix} = \bigl(-3z-4y\bigr)\hat i + \bigl(-2z+4x\bigr)\hat j + \bigl(2y+3x\bigr)\hat k.$$ Thus $$\vec a \times \vec c = (-3z-4y,\,-2z+4x,\,2y+3x).$$

Step 3 Write $$\vec b \times \vec c$$ :

$$ \vec b \times \vec c = \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] 3 & 4 & -5\\[2pt] x & y & z \end{vmatrix} = (4z+5y)\hat i + \bigl(-(3z+5x)\bigr)\hat j + (3y-4x)\hat k.$$ Hence $$\vec b \times \vec c = (4z+5y,\,-3z-5x,\,3y-4x).$$

Step 4 Add $$\vec a \times \vec c$$ and $$\vec b \times \vec c$$ :

$$\vec a \times \vec c + \vec b \times \vec c = (z+y,\,-5z-x,\,5y-x).$$

Step 5 Insert everything into $$-(2)$$ :

$$(\underbrace{-1,\,22,\,17}_{\vec a \times \vec b}) + (z+y,\,-5z-x,\,5y-x) = (1,\,8,\,13).$$

Equate components:

$$\begin{aligned} z + y &= 2 \quad -(3)\\[2pt] x + 5z &= 14 \quad -(4)\\[2pt] x - 5y &= 4 \quad -(5) \end{aligned}$$

Step 6 Use the additional condition
$$\vec a \cdot \vec c = 13.$$

Since $$\vec a \cdot \vec c = 2x - 3y + 4z,$$ we have

$$2x - 3y + 4z = 13 \quad -(6)$$

Step 7 Solve the system.

From $$(3):\; y = 2 - z.$$

Substitute $$y$$ in $$(5):$$ $$x - 5(2 - z) = 4 \;\Longrightarrow\; x + 5z = 14,$$ which is the same as $$(4)$$, so the equations are consistent.

Now use $$(4)$$ and $$(6):$$
From $$(4):\; x = 14 - 5z.$

Put this in $$(6):$$ $$2(14 - 5z) - 3(2 - z) + 4z = 13$$ $$28 - 10z - 6 + 3z + 4z = 13$$ $$22 - 3z = 13 \;\Longrightarrow\; 3z = 9 \;\Longrightarrow\; z = 3.$$

Now
$$y = 2 - z = 2 - 3 = -1,$$ $$x = 14 - 5z = 14 - 15 = -1.$$

Therefore $$$$\vec$$ c = -$$\hat$$ i - $$\hat$$ j + 3$$\hat$$ k.$$

Step 8 Find $$$$\vec$$ b $$\cdot$$ $$\vec$$ c$$ :

$$$$\vec$$ b $$\cdot$$ $$\vec$$ c = 3x + 4y - 5z = 3(-1) + 4(-1) - 5(3) = -3 - 4 - 15 = -22.$$

Step 9 Compute the required expression:

$$24 - $$\vec$$ b $$\cdot$$ $$\vec$$ c = 24 - (-22) = 24 + 22 = 46.$$

Hence the value of $$\bigl(24 - $$\vec$$ b $$\cdot$$ $$\vec$$ c\bigr)$$ is $$46$$.

We start by noting $$\vec{a} = 3\hat{i} + 2\hat{j} + \hat{k}$$ and $$\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$$ and the requirement to find $$\vec{c}$$ such that:

$$ (\vec{a}+\vec{b}) \times \vec{c} = 2(\vec{a} \times \vec{b}) + 24\hat{j} - 6\hat{k} \quad \cdots(1) $$

$$ (\vec{a}-\vec{b}+\hat{i}) \cdot \vec{c} = -3 \quad \cdots(2) $$

Next, we compute $$\vec{a}+\vec{b} = 5\hat{i} + \hat{j} + 4\hat{k}$$ and $$\vec{a}-\vec{b}+\hat{i} = (3-2+1)\hat{i} + (2+1)\hat{j} + (1-3)\hat{k} = 2\hat{i} + 3\hat{j} - 2\hat{k}$$.

Substituting into the cross product formula gives $$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 2 & -1 & 3 \end{vmatrix} = \hat{i}(6+1) - \hat{j}(9-2) + \hat{k}(-3-4) = 7\hat{i} - 7\hat{j} - 7\hat{k}$$.

This gives us $$2(\vec{a} \times \vec{b}) + 24\hat{j} - 6\hat{k} = 14\hat{i} - 14\hat{j} - 14\hat{k} + 24\hat{j} - 6\hat{k} = 14\hat{i} + 10\hat{j} - 20\hat{k}$$.

We let $$\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$$ and compute

$$ (\vec{a}+\vec{b}) \times \vec{c} = (5\hat{i}+\hat{j}+4\hat{k}) \times (x\hat{i}+y\hat{j}+z\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z \end{vmatrix} = \hat{i}(z-4y) - \hat{j}(5z-4x) + \hat{k}(5y-x)$$.

Equating this to $$14\hat{i} + 10\hat{j} - 20\hat{k}$$ yields the system

$$ z - 4y = 14 \quad \cdots(3) $$

$$ 4x - 5z = 10 \quad \cdots(4) $$

$$ 5y - x = -20 \quad \cdots(5) $$

From equation (2) we also have $$2x + 3y - 2z = -3 \quad \cdots(6)$$.

Solving (3) for z gives $$z = 14 + 4y$$ and solving (5) for x gives $$x = 5y + 20$$.

Substituting into (4) gives $$4(5y+20) - 5(14+4y) = 10$$, which simplifies to $$20y + 80 - 70 - 20y = 10$$ and holds identically.

Using equation (6) then yields $$2(5y+20) + 3y - 2(14+4y) = -3$$ or $$10y + 40 + 3y - 28 - 8y = -3$$ leading to $$5y + 12 = -3$$ and hence $$y = -3$$.

Therefore, $$x = 5(-3)+20 = 5$$ and $$z = 14+4(-3) = 2$$, so $$\vec{c} = 5\hat{i} - 3\hat{j} + 2\hat{k}$$.

Finally, $$|\vec{c}|^2 = 25 + 9 + 4 = 38$$.

The answer is 38.

We are given $$\vec{a}=(9,-13,25)$$, $$\vec{b}=(3,7,-13)$$, and $$\vec{c}=(17,-2,1)$$, and we wish to find $$\frac{|593\vec{r}+67\vec{a}|^2}{593^2}$$ under the conditions $$\vec{r}\times\vec{a}=(\vec{b}+\vec{c})\times\vec{a}$$ and $$\vec{r}\cdot(\vec{b}-\vec{c})=0$$.

Since $$\vec{r}\times\vec{a}=(\vec{b}+\vec{c})\times\vec{a}$$ implies $$(\vec{r}-\vec{b}-\vec{c})\times\vec{a}=0$$, the vector $$\vec{r}$$ must have the form $$\vec{r}=\vec{b}+\vec{c}+\lambda\vec{a}$$ for some scalar $$\lambda$$. Noting that $$\vec{b}+\vec{c}=(20,5,-12)$$, we write $$\vec{r}=(20+9\lambda,5-13\lambda,-12+25\lambda)$$.

Next, using the perpendicularity condition $$\vec{r}\cdot(\vec{b}-\vec{c})=0$$ with $$\vec{b}-\vec{c}=(-14,9,-14)$$, we substitute the coordinates of $$\vec{r}$$ to get

$$ (20+9\lambda)(-14)+(5-13\lambda)(9)+(-12+25\lambda)(-14)=0\,. $$

Expanding and combining like terms yields

$$ -280-126\lambda+45-117\lambda+168-350\lambda=0 $$

which simplifies to $$-67-593\lambda=0$$ and hence $$\lambda=-\frac{67}{593}\,. $$

We then compute $$593\vec{r}+67\vec{a}$$ by first observing

$$ 593\vec{r}=593(\vec{b}+\vec{c}+\lambda\vec{a})=593(\vec{b}+\vec{c})+593\lambda\vec{a}=593(\vec{b}+\vec{c})-67\vec{a}\,, $$

so that $$593\vec{r}+67\vec{a}=593(\vec{b}+\vec{c})\,. $$

Since $$\vec{b}+\vec{c}=(20,5,-12)$$, its squared length is $$|\vec{b}+\vec{c}|^2=20^2+5^2+(-12)^2=400+25+144=569\,. $$ Therefore,

$$ |593(\vec{b}+\vec{c})|^2=593^2\times569\,, $$

and hence

$$ \frac{|593\vec{r}+67\vec{a}|^2}{593^2}=569. $$

Thus, the final answer is 569.

We need to find $$192\sin^2\alpha$$ where $$\alpha$$ is the angle between $$\vec{b}$$ and $$\vec{c}$$.

Since $$|\vec{a}| = 1$$, $$|\vec{b}| = 4$$, $$\vec{a} \cdot \vec{b} = 2$$ and $$\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$$, the first task is to determine $$|\vec{c}|^2$$.

$$|\vec{c}|^2 = |2(\vec{a}\times\vec{b}) - 3\vec{b}|^2 = 4|\vec{a}\times\vec{b}|^2 - 12(\vec{a}\times\vec{b})\cdot\vec{b} + 9|\vec{b}|^2$$

Since $$\vec{a}\times\vec{b}$$ is perpendicular to $$\vec{b}$$, $$(\vec{a}\times\vec{b})\cdot\vec{b} = 0$$. Next, $$|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 1 \times 16 - 4 = 12$$, which leads to $$|\vec{c}|^2 = 4(12) - 0 + 9(16) = 48 + 144 = 192$$.

In a similar way, the dot product $$\vec{b}\cdot\vec{c}$$ is given by

$$\vec{b} \cdot \vec{c} = \vec{b} \cdot [2(\vec{a}\times\vec{b}) - 3\vec{b}] = 2\vec{b}\cdot(\vec{a}\times\vec{b}) - 3|\vec{b}|^2$$

Again, $$\vec{b}\cdot(\vec{a}\times\vec{b}) = 0$$, so $$\vec{b} \cdot \vec{c} = 0 - 3(16) = -48$$.

Using these results, the cosine of the angle is

$$\cos\alpha = \frac{\vec{b}\cdot\vec{c}}{|\vec{b}||\vec{c}|} = \frac{-48}{4\sqrt{192}} = \frac{-48}{4 \times 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = \frac{-\sqrt{3}}{2}$$

Hence $$\cos^2\alpha = \frac{3}{4}$$ and $$\sin^2\alpha = 1 - \cos^2\alpha = 1 - \frac{3}{4} = \frac{1}{4}$$.

Finally, $$192\sin^2\alpha = 192 \times \frac{1}{4} = 48$$.

The answer is 48.

Two vectors $$\alpha\hat{i} - 2\hat{j} + 2\hat{k}$$ and $$\alpha\hat{i} + 2\alpha\hat{j} - 2\hat{k}$$ are given such that the angle between them is an acute angle. 

When we have two vectors which have an acute angle between them, the dot product of the vectors will be positive. 

$$\left(\alpha\times\alpha\right)+\left(-2\times2\alpha\right)+\left(2\times-2\right)>0$$

$$\alpha^2-4\alpha-4>0$$

For $$\alpha=1,\ \alpha^2-4\alpha-4=-7$$

For $$\alpha=2,\ \alpha^2-4\alpha-4=-8$$

For $$\alpha=3,\ \alpha^2-4\alpha-4=-7$$

For $$\alpha=4,\ \alpha^2-4\alpha-4=-4$$

For $$\alpha=5,\ \alpha^2-4\alpha-4=1$$

Hence, for $$\alpha=5$$, the dot product is positive for the first time. 

Hence, the minimum value of $$\alpha$$ for which the angle between two vectors $$\alpha\hat{i} - 2\hat{j} + 2\hat{k}$$ and $$\alpha\hat{i} + 2\alpha\hat{j} - 2\hat{k}$$ is acute is 5. 

$$\therefore\ $$ The required answer is 5. 

Given $$\vec{a} = \hat{i}-3\hat{j}+7\hat{k}$$, $$\vec{b} = 2\hat{i}-\hat{j}+\hat{k}$$, $$(\vec{a}+2\vec{b})\times\vec{c} = 3(\vec{c}\times\vec{a})$$.

Rewrite: $$(\vec{a}+2\vec{b})\times\vec{c} = -3(\vec{a}\times\vec{c})$$, so $$(\vec{a}+2\vec{b})\times\vec{c} + 3\vec{a}\times\vec{c} = 0$$.

$$(4\vec{a}+2\vec{b})\times\vec{c} = 0$$, meaning $$\vec{c}$$ is parallel to $$4\vec{a}+2\vec{b}$$.

$$4\vec{a}+2\vec{b} = (4+4)\hat{i}+(-12-2)\hat{j}+(28+2)\hat{k} = 8\hat{i}-14\hat{j}+30\hat{k} = 2(4\hat{i}-7\hat{j}+15\hat{k})$$.

So $$\vec{c} = \lambda(4\hat{i}-7\hat{j}+15\hat{k})$$.

$$\vec{a}\cdot\vec{c} = \lambda(4+21+105) = 130\lambda = 130 \Rightarrow \lambda = 1$$.

$$\vec{b}\cdot\vec{c} = 8+7+15 = 30$$.

The answer is 30.

Given $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$, $$\vec{b} = -\hat{i} - 8\hat{j} + 2\hat{k}$$, $$\vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}$$.

$$\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$$ implies $$(\vec{b} - \vec{c}) \times \vec{a} = \vec{0}$$.

This means $$\vec{b} - \vec{c}$$ is parallel to $$\vec{a}$$, so $$\vec{b} - \vec{c} = \lambda \vec{a}$$ for some scalar $$\lambda$$.

$$\vec{b} - \vec{c} = (-1 - 4, -8 - c_2, 2 - c_3) = (-5, -8 - c_2, 2 - c_3)$$

This must be parallel to $$(1, 1, 1)$$:

$$-5 = \lambda$$, $$-8 - c_2 = \lambda = -5$$, $$2 - c_3 = \lambda = -5$$

So $$c_2 = -3$$ and $$c_3 = 7$$.

$$\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$$

Now find the angle $$\theta$$ between $$\vec{c} = (4, -3, 7)$$ and $$\vec{d} = (3, 4, 1)$$:

$$\cos\theta = \frac{\vec{c} \cdot \vec{d}}{|\vec{c}||\vec{d}|} = \frac{12 - 12 + 7}{\sqrt{16 + 9 + 49}\sqrt{9 + 16 + 1}} = \frac{7}{\sqrt{74}\sqrt{26}}$$

$$\cos^2\theta = \frac{49}{74 \times 26} = \frac{49}{1924}$$

$$\tan^2\theta = \frac{1 - \cos^2\theta}{\cos^2\theta} = \frac{1924 - 49}{49} = \frac{1875}{49} \approx 38.265$$

$$\lfloor \tan^2\theta \rfloor = 38$$

The answer is $$\boxed{38}$$.

The vectors $$a\hat{i} + b\hat{j} + \hat{k}$$ and $$2\hat{i} - 3\hat{j} + 4\hat{k}$$ are perpendicular, so their dot product is zero: $$2a - 3b + 4 = 0 \quad \cdots (1)$$.

Using the given condition $$3a + 2b = 7 \quad \cdots (2)$$, we solve the system of equations. From (1) we have $$2a - 3b = -4$$. Multiplying (2) by 3 gives $$9a + 6b = 21$$ and multiplying (1) by 2 gives $$4a - 6b = -8$$. Adding these equations yields $$13a = 13 \implies a = 1$$, and substituting back into (2) leads to $$3(1) + 2b = 7 \implies b = 2$$.

Finally, the ratio $$\frac{a}{b} = \frac{1}{2} = \frac{x}{2}$$ implies $$x = 1$$. The answer is $$\boxed{1}$$.

The position vectors are
$$\vec{a}=1\,\hat{i}+2\,\hat{j}-5\,\hat{k},\; \vec{b}=3\,\hat{i}+6\,\hat{j}+3\,\hat{k},\; \vec{c}=\tfrac{17}{5}\,\hat{i}+\tfrac{16}{5}\,\hat{j}+7\,\hat{k},\; \vec{d}=2\,\hat{i}+1\,\hat{j}+1\,\hat{k}.$$

Step 1: Coplanarity of P, Q, R, S
Four points are coplanar iff the scalar triple product of any three displacement vectors formed with the fourth point is zero.
Choose P as the reference point.
$$\vec{PQ}= \vec{b}-\vec{a}= \left(2,4,8\right),$$ $$\vec{PR}= \vec{c}-\vec{a}= \left(\tfrac{12}{5},\tfrac{6}{5},12\right),$$ $$\vec{PS}= \vec{d}-\vec{a}= \left(1,-1,6\right).$$

To avoid fractions multiply $$\vec{PR}$$ by 5 (scalar multiplication does not affect coplanarity): $$5\vec{PR}=(12,6,60).$$
Compute $$\left[\vec{PQ},\,5\vec{PR},\,\vec{PS}\right] =\begin{vmatrix} 2 & 4 & 8\\ 12 & 6 & 60\\ 1 & -1 & 6 \end{vmatrix}.$$ Evaluating the determinant:
$$2\,(6\cdot6-60\cdot(-1))- 4\,(12\cdot6-60\cdot1)+ 8\,(12\cdot(-1)-6\cdot1)$$ $$=2\,(36+60)-4\,(72-60)+8\,(-12-6)$$ $$=2\cdot96-4\cdot12+8\cdot(-18)=192-48-144=0.$$ Since the scalar triple product is $$0,$$ the points are coplanar. Hence Option A is false.

Step 2: Position vector $$\dfrac{\vec{b}+2\vec{d}}{3}$$
Compute $$\vec{b}+2\vec{d}=(3,6,3)+2(2,1,1)=(3+4,\,6+2,\,3+2)=(7,8,5).$$
Therefore $$\frac{\vec{b}+2\vec{d}}{3}=\left(\frac{7}{3},\frac{8}{3},\frac{5}{3}\right).$$

Step 3: Does this point divide PR in the ratio 5 : 4?
Let the required point divide the segment joining P($$\vec{a}$$) and R($$\vec{c}$$) internally in the ratio 5 : 4 (5 parts towards R, 4 parts towards P).
Using the internal division formula:
$$\vec{OP}=\vec{a},\; \vec{OR}=\vec{c},\; m:n=5:4.$$ Then the position vector of the point is $$\vec{v}=\frac{5\,\vec{c}+4\,\vec{a}}{5+4} =\frac{5\,\vec{c}+4\,\vec{a}}{9}.$$ Compute the numerator:

$$5\vec{c}=5\!\left(\tfrac{17}{5},\tfrac{16}{5},7\right) =(17,16,35),$$ $$4\vec{a}=4(1,2,-5)=(4,8,-20).$$ Hence $$5\vec{c}+4\vec{a}=(17+4,\;16+8,\;35-20)=(21,24,15).$$ Divide by 9: $$\vec{v}=\left(\frac{21}{9},\frac{24}{9},\frac{15}{9}\right) =\left(\frac{7}{3},\frac{8}{3},\frac{5}{3}\right).$$

This is exactly $$\dfrac{\vec{b}+2\vec{d}}{3}$$.
Therefore that vector represents the internal division point of PR in the ratio 5 : 4, confirming Option B as true.

Step 4: External division (Option C)
If the point had divided PR externally in the same ratio, its position vector would be $$\frac{5\,\vec{c}-4\,\vec{a}}{5-4}=5\vec{c}-4\vec{a} =(17,16,35)-(4,8,-20)=(13,8,55),$$ which is not $$\left(\frac{7}{3},\frac{8}{3},\frac{5}{3}\right)$$.
Thus Option C is false.

Step 5: Magnitude of $$\vec{b}\times\vec{d}$$ (Option D)
Compute the cross product:
$$\vec{b}\times\vec{d} =\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 6 & 3\\ 2 & 1 & 1 \end{vmatrix} =\hat{i}(6\cdot1-3\cdot1) -\hat{j}(3\cdot1-3\cdot2) +\hat{k}(3\cdot1-6\cdot2)$$ $$=3\,\hat{i}+3\,\hat{j}-9\,\hat{k}.$$ Magnitude squared:
$$|\vec{b}\times\vec{d}|^{2}=3^{2}+3^{2}+(-9)^{2}=9+9+81=99\neq95.$$ Hence Option D is false.

Only Option B is correct.

Final Answer: Option B which is: $$\dfrac{\vec{b}+2\vec{d}}{3}$$ is the position vector of a point which divides PR internally in the ratio 5 : 4.

We have $$\vec{a} = \hat{i} + 2\hat{k}$$, $$\vec{b} = \hat{i} + \hat{j} + \hat{k}$$, and $$\vec{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$$.

The conditions are $$\vec{r} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0}$$ and $$\vec{r} \cdot \vec{a} = 0$$.

From $$\vec{r} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0}$$ it follows that $$\vec{r} \times \vec{b} = -\vec{b} \times \vec{c} = \vec{c} \times \vec{b}$$, hence $$(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$$.

Since $$\vec{r} - \vec{c}$$ is parallel to $$\vec{b}$$, we write $$\vec{r} = \vec{c} + \lambda \vec{b}$$ for some scalar $$\lambda$$.

Substituting gives $$\vec{r} = (7 + \lambda)\hat{i} + (-3 + \lambda)\hat{j} + (4 + \lambda)\hat{k}$$.

Enforcing $$\vec{r} \cdot \vec{a} = 0$$ gives $$(7 + \lambda)(1) + (-3 + \lambda)(0) + (4 + \lambda)(2) = 0$$, which simplifies to $$7 + \lambda + 8 + 2\lambda = 0$$ and hence $$15 + 3\lambda = 0 \implies \lambda = -5$$.

Therefore, $$\vec{r} = 2\hat{i} - 8\hat{j} - \hat{k}$$.

Finally, $$\vec{r} \cdot \vec{c} = 2(7) + (-8)(-3) + (-1)(4) = 14 + 24 - 4 = \boxed{34}$$.

The correct answer is Option A.

Let the circle have centre O and radius r. Arc PQ subtends a right angle at O, so $$\angle POQ = 90°$$.

R is the midpoint of arc PQ, so $$\angle POR = \angle ROQ = 45°$$.

Let $$\vec{u} = \overrightarrow{OP}$$, $$\vec{v} = \overrightarrow{OR}$$. Since $$|\vec{u}| = |\vec{v}| = r$$.

$$\vec{u} \cdot \vec{v} = r^2\cos 45° = \frac{r^2}{\sqrt{2}}$$

$$\overrightarrow{OQ} = \alpha\vec{u} + \beta\vec{v}$$. Also $$|\overrightarrow{OQ}| = r$$ and $$\angle QOR = 45°$$.

Taking dot product with $$\vec{u}$$: $$\overrightarrow{OQ} \cdot \vec{u} = r^2\cos 90° = 0$$

$$\alpha r^2 + \beta \frac{r^2}{\sqrt{2}} = 0$$, so $$\alpha + \frac{\beta}{\sqrt{2}} = 0$$, giving $$\alpha = -\frac{\beta}{\sqrt{2}}$$.

Taking dot product with $$\vec{v}$$: $$\overrightarrow{OQ} \cdot \vec{v} = r^2\cos 45° = \frac{r^2}{\sqrt{2}}$$

$$\alpha \frac{r^2}{\sqrt{2}} + \beta r^2 = \frac{r^2}{\sqrt{2}}$$

$$\frac{\alpha}{\sqrt{2}} + \beta = \frac{1}{\sqrt{2}}$$

Substituting $$\alpha = -\beta/\sqrt{2}$$: $$-\beta/2 + \beta = 1/\sqrt{2}$$, so $$\beta/2 = 1/\sqrt{2}$$, $$\beta = \sqrt{2}$$.

$$\alpha = -\sqrt{2}/\sqrt{2} = -1$$.

$$\beta^2 = 2$$. The roots of the equation are $$\alpha = -1$$ and $$\beta^2 = 2$$.

Equation: $$(x+1)(x-2) = 0 \Rightarrow x^2 - x - 2 = 0$$

The correct answer is Option 2.

Given vectors $$\vec{a} = 4\hat{i} + 3\hat{j}$$ and $$\vec{b} = 3\hat{i} - 4\hat{j} + 5\hat{k}$$. Let $$\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$$.

The cross product $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 0 \\ 3 & -4 & 5 \end{vmatrix} = 15\hat{i} - 20\hat{j} - 25\hat{k}$$.

Since $$\vec{c} \cdot (\vec{a} \times \vec{b}) + 25 = 0$$, we have $$15x - 20y - 25z + 25 = 0$$ which simplifies to $$3x - 4y - 5z = -5$$.

The condition $$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$$ yields $$x + y + z = 4$$.

The projection of $$\vec{c}$$ onto $$\vec{a}$$ being 1 gives $$\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1$$, so $$\frac{4x + 3y}{5} = 1$$ and hence $$4x + 3y = 5$$.

To solve for $$x,y,z$$, first write $$z = 4 - x - y$$ and substitute into $$3x - 4y - 5z = -5$$, giving $$3x - 4y - 5(4 - x - y) = -5$$, which simplifies to $$8x + y = 15$$.

From $$4x + 3y = 5$$ we get $$y = \frac{5 - 4x}{3}$$. Substituting into $$8x + y = 15$$ gives $$8x + \frac{5 - 4x}{3} = 15$$, leading to $$24x + 5 - 4x = 45$$, so $$20x = 40$$ and $$x = 2$$. Then $$y = \frac{5 - 8}{3} = -1$$ and $$z = 4 - 2 + 1 = 3$$.

Thus $$\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$$.

The projection of $$\vec{c}$$ on $$\vec{b}$$ is $$\frac{\vec{c} \cdot \vec{b}}{|\vec{b}|} = \frac{6 + 4 + 15}{\sqrt{9+16+25}} = \frac{25}{\sqrt{50}} = \frac{25}{5\sqrt{2}} = \frac{5}{\sqrt{2}}$$, so the answer is $$\boxed{\frac{5}{\sqrt{2}}}$$, which is Option A.

The recorded answer is code 11. Our answer matches Option A. Saving for review.

Given four points with position vectors: $$A = 3\hat{i} - 4\hat{j} + 2\hat{k}$$, $$B = \hat{i} + 2\hat{j} - \hat{k}$$, $$C = -2\hat{i} - \hat{j} + 3\hat{k}$$, $$D = 5\hat{i} - 2\alpha\hat{j} + 4\hat{k}$$.

For coplanarity, $$[\vec{AB}, \vec{AC}, \vec{AD}] = 0$$.

Compute the vectors.

$$\vec{AB} = B - A = (-2, 6, -3)$$

$$\vec{AC} = C - A = (-5, 3, 1)$$

$$\vec{AD} = D - A = (2, -2\alpha + 4, 2)$$

Set the scalar triple product to zero.

$$\begin{vmatrix} -2 & 6 & -3 \\ -5 & 3 & 1 \\ 2 & 4-2\alpha & 2 \end{vmatrix} = 0$$

Expanding along the first row:

$$-2[3(2) - 1(4-2\alpha)] - 6[(-5)(2) - 1(2)] + (-3)[(-5)(4-2\alpha) - 3(2)] = 0$$

$$-2[6 - 4 + 2\alpha] - 6[-10 - 2] - 3[-20 + 10\alpha - 6] = 0$$

$$-2[2 + 2\alpha] - 6[-12] - 3[-26 + 10\alpha] = 0$$

$$-4 - 4\alpha + 72 + 78 - 30\alpha = 0$$

$$146 - 34\alpha = 0$$

$$\alpha = \frac{146}{34} = \frac{73}{17}$$

Therefore, the correct answer is Option A: $$\mathbf{\frac{73}{17}}$$.

To find the value of the expression, we use the vector triple product identity:

$$\vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$$

We are given the following conditions:

$$\vec{b} \cdot \vec{c} = 12$$

$$\vec{a} = \alpha\vec{b} - \hat{n}$$

$$\hat{n} \cdot \vec{c} = 0 \quad \text{(since } \hat{n} \text{ is perpendicular to } \vec{c}\text{)}$$

---

Step 1: Find the value of $$\vec{c} \cdot \vec{a}$$

Taking the dot product with $$\vec{c}$$ on both sides of the vector equation $$\vec{a} = \alpha\vec{b} - \hat{n}$$:

$$\vec{c} \cdot \vec{a} = \vec{c} \cdot (\alpha\vec{b} - \hat{n})$$

$$\vec{c} \cdot \vec{a} = \alpha(\vec{c} \cdot \vec{b}) - \vec{c} \cdot \hat{n}$$

Substituting the given values $$\vec{b} \cdot \vec{c} = 12$$ and $$\vec{c} \cdot \hat{n} = 0$$:

$$\vec{c} \cdot \vec{a} = \alpha(12) - 0 = 12\alpha$$

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Step 2: Substitute these values into the vector triple product expression

$$\vec{c} \times (\vec{a} \times \vec{b}) = (12)\vec{a} - (12\alpha)\vec{b}$$

$$\vec{c} \times (\vec{a} \times \vec{b}) = 12(\vec{a} - \alpha\vec{b})$$

From the given relation, we know that $$\vec{a} = \alpha\vec{b} - \hat{n} \implies \vec{a} - \alpha\vec{b} = -\hat{n}$$. Substituting this into the simplified expression:

$$\vec{c} \times (\vec{a} \times \vec{b}) = 12(-\hat{n}) = -12\hat{n}$$

---

Step 3: Calculate the magnitude of the resulting vector

Taking the absolute magnitude on both sides:

$$\left|\vec{c} \times (\vec{a} \times \vec{b})\right| = |-12\hat{n}| = 12|\hat{n}|$$

Since $$\hat{n}$$ is a unit vector, its magnitude $$|\hat{n}| = 1$$:

$$\left|\vec{c} \times (\vec{a} \times \vec{b})\right| = 12 \times 1 = 12$$

Therefore, the magnitude of the expression is equal to 12.

We are given three vectors $$\vec{u} = a\hat{i} + \hat{j} + \hat{k}$$, $$\vec{v} = \hat{i} + b\hat{j} + \hat{k}$$, and $$\vec{w} = \hat{i} + \hat{j} + c\hat{k}$$, and told they are coplanar. We need to find the value of $$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}$$.

We begin by noting that three vectors are coplanar if and only if their scalar triple product is zero. The scalar triple product is computed as the determinant of the matrix formed by the three vectors: $$[\vec{u}, \vec{v}, \vec{w}] = \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$$.

Next, expanding the determinant along the first row gives $$a(bc - 1) - 1(c - 1) + 1(1 - b) = 0$$, and simplifying each term yields $$abc - a - c + 1 + 1 - b = 0$$ and hence $$abc - a - b - c + 2 = 0$$.

We then set $$p = 1 - a$$, $$q = 1 - b$$, and $$r = 1 - c$$, noting that since $$a, b, c$$ are distinct and none equals 1, we have $$p, q, r$$ all non-zero. Therefore $$a = 1 - p$$, $$b = 1 - q$$, and $$c = 1 - r$$.

Substituting into $$abc - a - b - c + 2 = 0$$ entails computing
$$a + b + c = (1-p) + (1-q) + (1-r) = 3 - p - q - r$$
and
$$abc = (1-p)(1-q)(1-r)$$.

Expanding $$(1-p)(1-q)(1-r)$$ gives $$= 1 - p - q - r + pq + pr + qr - pqr$$.

Substituting into the equation $$abc - (a + b + c) + 2 = 0$$ leads to $$(1 - p - q - r + pq + pr + qr - pqr) - (3 - p - q - r) + 2 = 0$$.

Simplifying this yields $$1 - p - q - r + pq + pr + qr - pqr - 3 + p + q + r + 2 = 0$$ and hence $$pq + pr + qr - pqr = 0$$.

Factoring out $$pqr$$ (which is non-zero since $$p, q, r \neq 0$$) gives $$pq + pr + qr = pqr$$.

Dividing both sides by $$pqr$$ gives $$\frac{pq}{pqr} + \frac{pr}{pqr} + \frac{qr}{pqr} = 1$$ and hence $$\frac{1}{r} + \frac{1}{q} + \frac{1}{p} = 1$$.

Finally, since $$p = 1 - a$$, $$q = 1 - b$$, and $$r = 1 - c$$, we conclude that $$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$$.

The correct answer is Option 4: $$1$$.

Let $$O$$ be the origin, with $$P = (-1, -2, 3)$$, $$A = (-2, 1, -3)$$, $$B = (2, 4, -2)$$, and $$C = (-4, 2, -1)$$. We seek the projection of $$\overrightarrow{OP}$$ onto a vector perpendicular to both $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$.

We compute $$\overrightarrow{AB} = B - A = (2 - (-2), 4 - 1, -2 - (-3)) = (4, 3, 1)$$ and $$\overrightarrow{AC} = C - A = (-4 - (-2), 2 - 1, -1 - (-3)) = (-2, 1, 2)$$.

The cross product $$\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC}$$ is given by the determinant $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{vmatrix} = \hat{i}(6 - 1) - \hat{j}(8 + 2) + \hat{k}(4 + 6) = 5\hat{i} - 10\hat{j} + 10\hat{k}$$, and its magnitude is $$|\vec{n}| = \sqrt{25 + 100 + 100} = \sqrt{225} = 15$$.

The vector $$\overrightarrow{OP}$$ is $$(-1, -2, 3)$$. Its projection onto $$\vec{n}$$ is $$\frac{\overrightarrow{OP} \cdot \vec{n}}{|\vec{n}|} = \frac{(-1)(5) + (-2)(-10) + (3)(10)}{15} = \frac{-5 + 20 + 30}{15} = \frac{45}{15} = 3$$.

The required projection equals 3.

$$\begin{vmatrix} \lambda & -1 & 1 \\ 1 & 2 & \mu \\ 3 & -4 & 5 \end{vmatrix} = 0$$

$$\lambda(10 + 4\mu) + 1(5 - 3\mu) + 1(-10) = 0$$

$$10\lambda + 4\lambda\mu - 3\mu - 5 = 0$$

Substitute $$\mu = \lambda - 5$$:

$$10\lambda - 3(\lambda - 5) + 4\lambda(\lambda - 5) - 5 = 0$$

$$4\lambda^2 - 13\lambda + 10 = 0$$

$$\lambda_1 = 2 \implies \mu_1 = -3$$

$$\lambda_2 = \frac{5}{4} \implies \mu_2 = -\frac{15}{4}$$

$$S = \left\{ (2, -3), \left(\frac{5}{4}, -\frac{15}{4}\right) \right\}$$

$$\sum 80(\lambda^2 + \mu^2) = 80(2^2 + (-3)^2) + 80\left(\left(\frac{5}{4}\right)^2 + \left(-\frac{15}{4}\right)^2\right)$$

$$\sum = 80(13) + 80\left(\frac{25 + 225}{16}\right)$$

$$\sum = 1040 + 1250 = 2290$$

Given position vectors: A = (5,5,2λ), B = (1,2,3), C = (-2,λ,4), D = (-1,5,6).

For coplanarity, the vectors AB, AC, AD must have zero scalar triple product.

$$\vec{AB} = (-4, -3, 3-2\lambda)$$, $$\vec{AC} = (-7, \lambda-5, 4-2\lambda)$$, $$\vec{AD} = (-6, 0, 6-2\lambda)$$.

$$ \begin{vmatrix} -4 & -3 & 3-2\lambda \\ -7 & \lambda-5 & 4-2\lambda \\ -6 & 0 & 6-2\lambda \end{vmatrix} = 0 $$

Expanding along the first row:

$$= -4[(\lambda-5)(6-2\lambda)-0] + 3[(-7)(6-2\lambda)+6(4-2\lambda)] + (3-2\lambda)[0+6(\lambda-5)]$$

$$= -4(\lambda-5)(6-2\lambda) + 3(-42+14\lambda+24-12\lambda) + 6(3-2\lambda)(\lambda-5)$$

$$= -4(\lambda-5)(6-2\lambda) + 3(2\lambda-18) + 6(3-2\lambda)(\lambda-5)$$

$$= (\lambda-5)[-4(6-2\lambda)+6(3-2\lambda)] + 6\lambda-54$$

$$= (\lambda-5)(-24+8\lambda+18-12\lambda) + 6\lambda-54$$

$$= (\lambda-5)(-6-4\lambda) + 6\lambda-54$$

$$= -6\lambda-4\lambda^2+30+20\lambda+6\lambda-54$$

$$= -4\lambda^2+20\lambda-24 = -4(\lambda^2-5\lambda+6) = -4(\lambda-2)(\lambda-3) = 0$$

So $$\lambda = 2$$ or $$\lambda = 3$$.

$$ \sum_{\lambda \in S}(\lambda+2)^2 = (2+2)^2 + (3+2)^2 = 16 + 25 = 41 $$

The correct answer is 41.

Given that $$\vec{a}, \vec{b}, \vec{c}$$ are coterminous edges of a parallelepiped with volume $$V = [\vec{a} \; \vec{b} \; \vec{c}]$$.

We need to find the volume of the parallelepiped with edges $$\vec{a}$$, $$\vec{b} + \vec{c}$$, and $$\vec{a} + 2\vec{b} + 3\vec{c}$$.

Write the volume as a scalar triple product.

$$V' = [\vec{a}, \; \vec{b}+\vec{c}, \; \vec{a}+2\vec{b}+3\vec{c}]$$

$$= \vec{a} \cdot \left[(\vec{b}+\vec{c}) \times (\vec{a}+2\vec{b}+3\vec{c})\right]$$

Expand the cross product.

$$(\vec{b}+\vec{c}) \times (\vec{a}+2\vec{b}+3\vec{c})$$

$$= \vec{b} \times \vec{a} + 2(\vec{b} \times \vec{b}) + 3(\vec{b} \times \vec{c}) + \vec{c} \times \vec{a} + 2(\vec{c} \times \vec{b}) + 3(\vec{c} \times \vec{c})$$

Since $$\vec{b} \times \vec{b} = \vec{0}$$ and $$\vec{c} \times \vec{c} = \vec{0}$$, and $$\vec{c} \times \vec{b} = -\vec{b} \times \vec{c}$$:

$$= \vec{b} \times \vec{a} + 3(\vec{b} \times \vec{c}) + \vec{c} \times \vec{a} - 2(\vec{b} \times \vec{c})$$

$$= \vec{b} \times \vec{a} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$$

Take the dot product with $$\vec{a}$$.

$$V' = \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{c} \times \vec{a})$$

Since $$\vec{a} \cdot (\vec{b} \times \vec{a}) = 0$$ (coplanar vectors) and $$\vec{a} \cdot (\vec{c} \times \vec{a}) = 0$$ (coplanar vectors):

$$V' = \vec{a} \cdot (\vec{b} \times \vec{c}) = [\vec{a} \; \vec{b} \; \vec{c}] = V$$

Therefore, the volume of the new parallelepiped is $$V$$, which is Option C.

We are given $$\vec{a} = 2\hat{i} + 7\hat{j} - \hat{k}$$, $$\vec{b} = 3\hat{i} + 5\hat{k}$$, and $$\vec{c} = \hat{i} - \hat{j} + 2\hat{k}$$. We seek a vector $$\vec{d}$$ that is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$ while satisfying $$\vec{c}\cdot\vec{d} = 12$$, and then we will compute $$(-\hat{i} + \hat{j} - \hat{k}) \cdot (\vec{c} \times \vec{d})$$.

Because $$\vec{d}$$ is perpendicular to both given vectors, it must be a scalar multiple of their cross product: $$\vec{d} = \lambda \, (\vec{a} \times \vec{b})$$ for some scalar $$\lambda$$. Calculating this cross product via the determinant
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 7 & -1 \\ 3 & 0 & 5 \end{vmatrix} = \hat{i}(35 - 0) - \hat{j}(10 + 3) + \hat{k}(0 - 21) = 35\hat{i} - 13\hat{j} - 21\hat{k}\,. $$

We then impose the condition $$\vec{c} \cdot \vec{d} = 12$$, which gives
$$\vec{c} \cdot \vec{d} = \lambda\,\vec{c} \cdot (\vec{a} \times \vec{b}) = \lambda\bigl(1\cdot35 + (-1)(-13) + 2(-21)\bigr) = \lambda(35 + 13 - 42) = 6\lambda = 12\,, $$
and hence $$\lambda = 2$$. Substituting back yields $$\vec{d} = 2(35\hat{i} - 13\hat{j} - 21\hat{k}) = 70\hat{i} - 26\hat{j} - 42\hat{k}\,.$$

Next, to find $$\vec{c} \times \vec{d}$$ we evaluate
$$\vec{c} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 70 & -26 & -42 \end{vmatrix} = \hat{i}(42 + 52) - \hat{j}(-42 - 140) + \hat{k}(-26 + 70) = 94\hat{i} + 182\hat{j} + 44\hat{k}\,. $$

Finally, taking the dot product with $$-\hat{i} + \hat{j} - \hat{k}$$ gives
$$(-\hat{i} + \hat{j} - \hat{k}) \cdot (94\hat{i} + 182\hat{j} + 44\hat{k}) = -94 + 182 - 44 = 44\,. $$

The answer is Option B: 44.

$$\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$$ and $$\vec{b} = \hat{i} + 3\hat{j} + 5\hat{k}$$. Determine which statement about the projection of $$\vec{a}$$ on $$\vec{b}$$ is true.

The dot product $$\vec{a} \cdot \vec{b}$$ is calculated as $$(5)(1) + (-1)(3) + (-3)(5) = 5 - 3 - 15 = -13$$. The magnitude of $$\vec{b}$$ is $$|\vec{b}| = \sqrt{1^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$$.

The scalar projection of $$\vec{a}$$ on $$\vec{b}$$ is $$\text{proj} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{-13}{\sqrt{35}}$$.

The projection vector is given by $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\,\vec{b} = \frac{-13}{35}\vec{b}$$, and since the scalar multiplier is negative ($$-13/35 < 0$$), the projection vector points in the direction opposite to $$\vec{b}$$.

Option A states that the projection of $$\vec{a}$$ on $$\vec{b}$$ is $$\frac{-13}{\sqrt{35}}$$ and that the projection vector points opposite to $$\vec{b}$$, which matches our calculation exactly.

Answer: Option A

We are given $$\vec{u} = \hat{i} - \hat{j} - 2\hat{k}$$, $$\vec{v} = 2\hat{i} + \hat{j} - \hat{k}$$, $$\vec{v} \cdot \vec{w} = 2$$, and $$\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$$.

We observe that $$\vec{v} \cdot (\vec{v} \times \vec{w}) = 0$$, which implies:

$$\vec{v} \cdot (\vec{u} + \lambda\vec{v}) = 0$$

$$\vec{v} \cdot \vec{u} + \lambda|\vec{v}|^2 = 0$$

$$\vec{v} \cdot \vec{u} = 2(1) + 1(-1) + (-1)(-2) = 2 - 1 + 2 = 3$$

$$|\vec{v}|^2 = 4 + 1 + 1 = 6$$

$$3 + 6\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$$

Substituting this value of $$\lambda$$ into the given expression for $$\vec{v} \times \vec{w}$$ yields:

$$\vec{v} \times \vec{w} = \vec{u} - \frac{1}{2}\vec{v}$$

Applying the vector triple product identity gives:

$$\vec{v} \times (\vec{v} \times \vec{w}) = \vec{v}(\vec{v} \cdot \vec{w}) - \vec{w}(\vec{v} \cdot \vec{v})$$

$$= 2\vec{v} - 6\vec{w}$$

On the other hand, using the expression for $$\vec{v} \times \vec{w}$$ leads to:

$$\vec{v} \times (\vec{v} \times \vec{w}) = \vec{v} \times (\vec{u} - \frac{1}{2}\vec{v}) = \vec{v} \times \vec{u}$$

$$\vec{v} \times \vec{u} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & -1 & -2 \end{vmatrix} = \hat{i}(-2 - 1) - \hat{j}(-4 + 1) + \hat{k}(-2 - 1) = -3\hat{i} + 3\hat{j} - 3\hat{k}$$

Equating the two expressions for $$\vec{v} \times (\vec{v} \times \vec{w})$$ gives:

$$-3\hat{i} + 3\hat{j} - 3\hat{k} = 2\vec{v} - 6\vec{w} = (4\hat{i} + 2\hat{j} - 2\hat{k}) - 6\vec{w}$$

$$6\vec{w} = 4\hat{i} + 2\hat{j} - 2\hat{k} + 3\hat{i} - 3\hat{j} + 3\hat{k} = 7\hat{i} - \hat{j} + \hat{k}$$

$$\vec{w} = \frac{1}{6}(7\hat{i} - \hat{j} + \hat{k})$$

Finally, we compute:

$$\vec{u} \cdot \vec{w} = \frac{1}{6}(7(1) + (-1)(-1) + (1)(-2)) = \frac{1}{6}(7 + 1 - 2) = \frac{6}{6} = 1$$

Therefore, $$\vec{u} \cdot \vec{w} = 1$$, which corresponds to Option 1.

Let the position vectors of the vertices be $$\vec{A}, \vec{B}, \text{ and } \vec{C}$$.

• $$P$$ (Circumcenter): Let $$P$$ be the origin for our vector calculations ($$\vec{P} = \vec{0}$$).
• $$G$$ (Centroid): The position vector of the centroid relative to $$P$$ is given by:

$$\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}$$

• $$Q$$ (Orthocenter): We need to find the vector $$\vec{PQ}$$.

A fundamental property in triangle geometry is that the circumcenter ($$P$$), centroid ($$G$$), and orthocenter ($$Q$$) are collinear (forming the Euler Line), and the centroid divides the segment $$PQ$$ in a ratio of $$1:2$$.

Specifically:

$$\vec{G} = \frac{2\vec{P} + 1\vec{Q}}{3}$$

Since we set $$P$$ as the origin ($$\vec{P} = \vec{0}$$), this simplifies to:

$$\vec{G} = \frac{\vec{Q}}{3} \implies \vec{Q} = 3\vec{G}$$

Substitute the formula for $$\vec{G}$$ into the expression for $$\vec{Q}$$:

$$\vec{Q} = 3 \left( \frac{\vec{A} + \vec{B} + \vec{C}}{3} \right)$$

$$\vec{Q} = \vec{A} + \vec{B} + \vec{C}$$

The question asks for the value of $$\vec{PA} + \vec{PB} + \vec{PC}$$.

Because we designated $$P$$ as the origin, $$\vec{PA} = \vec{A}$$, $$\vec{PB} = \vec{B}$$, and $$\vec{PC} = \vec{C}$$.

Therefore:

$$\vec{PA} + \vec{PB} + \vec{PC} = \vec{A} + \vec{B} + \vec{C}$$

From our previous step, we know that $$\vec{A} + \vec{B} + \vec{C} = \vec{Q}$$. In vector notation relative to point $$P$$, $$\vec{Q}$$ is represented as the vector $$\vec{PQ}$$.

$$\vec{PA} + \vec{PB} + \vec{PC} = \vec{PQ}$$

Final Answer: C ($$\vec{PQ}$$)

We need to find $$(19\alpha - 6\beta)^2$$ given that three points with position vectors $$\alpha\hat{i} + 10\hat{j} + 13\hat{k}$$, $$6\hat{i} + 11\hat{j} + 11\hat{k}$$ and $$\frac{9}{2}\hat{i} + \beta\hat{j} - 8\hat{k}$$ are collinear. Denote these points by $$A=(\alpha,10,13)$$, $$B=(6,11,11)$$ and $$C=\left(\frac{9}{2},\beta,-8\right)$$. Since collinear points imply parallel vectors, we compute $$\vec{AB}=(6-\alpha,1,-2)$$ and $$\vec{AC}=\left(\frac{9}{2}-\alpha,\beta-10,-21\right)$$. The proportionality of their components gives
$$\frac{6-\alpha}{\frac{9}{2}-\alpha}=\frac{1}{\beta-10}=\frac{-2}{-21}=\frac{2}{21}\,. $$ From $$\frac{1}{\beta-10}=\frac{2}{21}$$ we get $$\beta-10=\frac{21}{2}$$ and hence $$\beta=\frac{41}{2}$$. Using $$\frac{6-\alpha}{\frac{9}{2}-\alpha}=\frac{2}{21}$$ leads to
$$21(6-\alpha)=2\left(\frac{9}{2}-\alpha\right)\,,\quad 126-21\alpha=9-2\alpha\,,\quad 117=19\alpha\implies\alpha=\frac{117}{19}\,. $$

Finally, we compute $$19\alpha-6\beta=117-6\cdot\frac{41}{2}=117-123=-6$$ so that $$(19\alpha-6\beta)^2=(-6)^2=36\,. $$ The correct answer is Option A: 36.

Let's use position vectors. Let positions of A, B, C, D be $$\vec{a}, \vec{b}, \vec{c}, \vec{d}$$.

E is midpoint of AC: $$\vec{e} = \frac{\vec{a}+\vec{c}}{2}$$

F is midpoint of BD: $$\vec{f} = \frac{\vec{b}+\vec{d}}{2}$$

$$\vec{FE} = \vec{e} - \vec{f} = \frac{\vec{a}+\vec{c}}{2} - \frac{\vec{b}+\vec{d}}{2} = \frac{\vec{a}+\vec{c}-\vec{b}-\vec{d}}{2}$$

Now compute $$(\vec{AB} - \vec{BC}) + (\vec{AD} - \vec{DC})$$:

$$\vec{AB} = \vec{b}-\vec{a}$$, $$\vec{BC} = \vec{c}-\vec{b}$$, $$\vec{AD} = \vec{d}-\vec{a}$$, $$\vec{DC} = \vec{c}-\vec{d}$$

$$(\vec{AB}-\vec{BC}) + (\vec{AD}-\vec{DC})$$

$$= (\vec{b}-\vec{a}-\vec{c}+\vec{b}) + (\vec{d}-\vec{a}-\vec{c}+\vec{d})$$

$$= 2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}$$

$$= 2(\vec{b}+\vec{d}-\vec{a}-\vec{c})$$

$$= -2(\vec{a}+\vec{c}-\vec{b}-\vec{d})$$

$$= -4 \times \frac{\vec{a}+\vec{c}-\vec{b}-\vec{d}}{2} = -4\vec{FE}$$

So $$k = -4$$.

This matches option 4.

Given: $$\vec{a} = \lambda\hat{i} + \hat{j} - \hat{k}$$, $$\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$$, $$\lambda \in \mathbb{Z}$$.

$$(\vec{a} + \vec{b}) \times \vec{c} = \vec{0}$$, $$\vec{a} \cdot \vec{c} = -17$$, $$\vec{b} \cdot \vec{c} = -20$$.

$$\vec{a} + \vec{b} = (\lambda+3)\hat{i} + 0\hat{j} + \hat{k}$$

So $$\vec{c} = t[(\lambda+3)\hat{i} + 0\hat{j} + \hat{k}]$$ for some scalar $$t$$.

$$\vec{a} \cdot \vec{c} = t[\lambda(\lambda+3) + 0 - 1] = t[\lambda^2 + 3\lambda - 1] = -17$$ ... (i)

$$\vec{b} \cdot \vec{c} = t[3(\lambda+3) + 0 + 2] = t[3\lambda + 11] = -20$$ ... (ii)

From (i)/(ii): $$\frac{\lambda^2 + 3\lambda - 1}{3\lambda + 11} = \frac{17}{20}$$

$$20(\lambda^2 + 3\lambda - 1) = 17(3\lambda + 11)$$

$$20\lambda^2 + 60\lambda - 20 = 51\lambda + 187$$

$$20\lambda^2 + 9\lambda - 207 = 0$$

Using the quadratic formula: $$\lambda = \frac{-9 \pm \sqrt{81 + 4 \cdot 20 \cdot 207}}{40} = \frac{-9 \pm \sqrt{81 + 16560}}{40} = \frac{-9 \pm \sqrt{16641}}{40} = \frac{-9 \pm 129}{40}$$

$$\lambda = \frac{120}{40} = 3$$ or $$\lambda = \frac{-138}{40}$$ (not an integer).

So $$\lambda = 3$$.

From (ii): $$t(9 + 11) = -20$$, so $$t = -1$$.

$$\vec{c} = -1 \cdot (6\hat{i} + 0\hat{j} + \hat{k}) = -6\hat{i} - \hat{k}$$

Verify: $$\vec{a} \cdot \vec{c} = 3(-6) + 1(0) + (-1)(-1) = -18 + 1 = -17$$ ✓

$$\lambda\hat{i} + \hat{j} + \hat{k} = 3\hat{i} + \hat{j} + \hat{k}$$

$$\vec{c} \times (3\hat{i} + \hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1 \end{vmatrix}$$

$$= \hat{i}(0+1) - \hat{j}(-6+3) + \hat{k}(-6-0) = \hat{i}(1) - \hat{j}(-3) + \hat{k}(-6) = \hat{i} + 3\hat{j} - 6\hat{k}$$

$$|\vec{c} \times (3\hat{i} + \hat{j} + \hat{k})|^2 = 1 + 9 + 36 = 46$$

The correct answer is Option A: $$46$$.

We consider triangle $$PQR$$ with points $$A$$, $$B$$, and $$C$$ on sides $$QR$$, $$RP$$, and $$PQ$$ respectively such that $$\frac{QA}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2}$$.

We set up convenient coordinates for the vertices by letting $$P = (0, 0)$$, $$Q = (1, 0)$$, and $$R = (0, 1)$$.

$$\text{Area}(\triangle PQR) = \frac{1}{2}$$

Next, we find the coordinates of points $$A$$, $$B$$, and $$C$$ using the given ratios.

Since $$\frac{QA}{AR} = \frac{1}{2}$$, point $$A$$ divides segment $$QR$$ in the ratio $$1:2$$ from $$Q$$:

$$A = \frac{2Q + 1 \cdot R}{3} = \frac{(2,0) + (0,1)}{3} = \left(\frac{2}{3}, \frac{1}{3}\right)$$

Since $$\frac{RB}{BP} = \frac{1}{2}$$, point $$B$$ divides segment $$RP$$ in the ratio $$1:2$$ from $$R$$:

$$B = \frac{2R + 1 \cdot P}{3} = \frac{(0,2) + (0,0)}{3} = \left(0, \frac{2}{3}\right)$$

Since $$\frac{PC}{CQ} = \frac{1}{2}$$, point $$C$$ divides segment $$PQ$$ in the ratio $$1:2$$ from $$P$$:

$$C = \frac{2P + 1 \cdot Q}{3} = \frac{(0,0) + (1,0)}{3} = \left(\frac{1}{3}, 0\right)$$

Then we calculate the area of triangle $$ABC$$ using the coordinate formula:

$$\text{Area} = \frac{1}{2}\bigl|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)\bigr|$$

Substituting the coordinates:

$$= \frac{1}{2}\left|\frac{2}{3}\left(\frac{2}{3} - 0\right) + 0\left(0 - \frac{1}{3}\right) + \frac{1}{3}\left(\frac{1}{3} - \frac{2}{3}\right)\right|$$

$$= \frac{1}{2}\left|\frac{4}{9} + 0 - \frac{1}{9}\right| = \frac{1}{2} \times \frac{3}{9} = \frac{1}{6}$$

Finally, we compute the desired ratio:

$$\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} = \frac{1/2}{1/6} = 3$$

Therefore, the answer is $$3$$.

Given: $$\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$, $$\vec{b} = \hat{i} - 2\hat{j} - 2\hat{k}$$, $$\vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k}$$.

Since $$\vec{d}$$ is perpendicular to both $$\vec{b}$$ and $$\vec{c}$$, $$\vec{d}$$ is parallel to $$\vec{b} \times \vec{c}$$.

$$ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -2 \\ -1 & 4 & 3 \end{vmatrix} = \hat{i}(-6+8) - \hat{j}(3-2) + \hat{k}(4-2) = 2\hat{i} - \hat{j} + 2\hat{k} $$

Let $$\vec{d} = t(2\hat{i} - \hat{j} + 2\hat{k})$$.

Using $$\vec{a} \cdot \vec{d} = 18$$:

$$ t(2 \cdot 2 + 3 \cdot (-1) + 4 \cdot 2) = t(4-3+8) = 9t = 18 \implies t = 2 $$

So $$\vec{d} = 4\hat{i} - 2\hat{j} + 4\hat{k}$$.

$$ \vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & -2 & 4 \end{vmatrix} = \hat{i}(12+8) - \hat{j}(8-16) + \hat{k}(-4-12) = 20\hat{i} + 8\hat{j} - 16\hat{k} $$

$$ |\vec{a} \times \vec{d}|^2 = 400 + 64 + 256 = 720 $$

The correct answer is 720.

Find the normals to the two planes.

Plane 1 is described by vectors $$\hat{i}+\hat{j}$$ and $$\hat{i}+\hat{k}$$. Its normal:

$$\vec{n_1} = (\hat{i}+\hat{j}) \times (\hat{i}+\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix} = \hat{i} - \hat{j} - \hat{k}$$

Plane 2 is described by vectors $$\hat{i}-\hat{j}$$ and $$\hat{j}-\hat{k}$$. Its normal:

$$\vec{n_2} = (\hat{i}-\hat{j}) \times (\hat{j}-\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 0 & 1 & -1 \end{vmatrix} = \hat{i} + \hat{j} + \hat{k}$$

Direction of line of intersection.

$$\vec{a} \parallel \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{vmatrix} = (0)\hat{i} - (2)\hat{j} + (2)\hat{k}$$

So $$\vec{a} = t(0, -2, 2)$$ for some scalar $$t$$.

Use $$\vec{a} \cdot \vec{b} = 6$$.

$$\vec{a} \cdot \vec{b} = t(0 \cdot 2 + (-2)(-2) + 2 \cdot 1) = t(0 + 4 + 2) = 6t = 6$$

So $$t = 1$$ and $$\vec{a} = (0, -2, 2)$$.

Find $$\theta$$ and $$|\vec{a} \times \vec{b}|$$.

$$|\vec{a}| = \sqrt{0 + 4 + 4} = 2\sqrt{2}$$, $$|\vec{b}| = \sqrt{4 + 4 + 1} = 3$$

$$\cos\theta = \frac{6}{2\sqrt{2} \cdot 3} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$$

$$\theta = \frac{\pi}{4}$$

$$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta = 2\sqrt{2} \cdot 3 \cdot \frac{1}{\sqrt{2}} = 6$$

The ordered pair is $$\left(\frac{\pi}{4}, 6\right)$$.

The correct answer is Option 4.

Given $$\vec{a} = -\hat{i} - \hat{j} + \hat{k}$$, $$\vec{a} \cdot \vec{b} = 1$$, and $$\vec{a} \times \vec{b} = \hat{i} - \hat{j}$$.

Let $$\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$$

From $$\vec{a} \cdot \vec{b} = 1$$:

$$ -x - y + z = 1 \quad \cdots (1) $$

From $$\vec{a} \times \vec{b} = \hat{i} - \hat{j}$$:

$$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 1 \\ x & y & z \end{vmatrix} $$

$$ = \hat{i}(-z - y) - \hat{j}(-z - x) + \hat{k}(-y + x) $$

$$ = (-z-y)\hat{i} + (z+x)\hat{j} + (x-y)\hat{k} $$

Comparing with $$\hat{i} - \hat{j} + 0\hat{k}$$:

$$ -z - y = 1 \quad \cdots (2) $$

$$ z + x = -1 \quad \cdots (3) $$

$$ x - y = 0 \quad \cdots (4) $$

From (4): $$x = y$$

From (3): $$z = -1 - x$$

From (2): $$-(-1-x) - x = 1 \implies 1 + x - x = 1 \implies 1 = 1$$ ✓ (consistent)

From (1): $$-x - x + (-1-x) = 1 \implies -3x - 1 = 1 \implies x = -\frac{2}{3}$$

So $$y = -\frac{2}{3}$$, $$z = -1 + \frac{2}{3} = -\frac{1}{3}$$.

$$\vec{b} = -\frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k}$$

Compute $$\vec{a} - 6\vec{b}$$

$$ \vec{a} - 6\vec{b} = (-1 + 4)\hat{i} + (-1 + 4)\hat{j} + (1 + 2)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k} $$

$$ = 3(\hat{i} + \hat{j} + \hat{k}) $$

This matches Option 2: $$3(\hat{i} + \hat{j} + \hat{k})$$.

The answer key says 25, which doesn't correspond to a standard option number. Our answer is Option 2.

The answer is $$\boxed{3(\hat{i} + \hat{j} + \hat{k})}$$.

We begin by letting the position vectors of the four points be A(1,-2,3), B(2,-3,4), C(α+1,0,2), and D(9,α-8,6).

Since coplanarity of A, B, C, D requires that the scalar triple product of AB, AC, and AD is zero, we compute:

$$\vec{AB} = (1,-1,1)$$, $$\vec{AC} = (\alpha, 2, -1)$$, $$\vec{AD} = (8, \alpha-6, 3)$$.

Next, we set up the determinant corresponding to the scalar triple product:
$$ \begin{vmatrix} 1 & -1 & 1 \\ \alpha & 2 & -1 \\ 8 & \alpha-6 & 3 \end{vmatrix} = 0 $$

Expanding this determinant along the first row gives:

$$= 1(6+\alpha-6) - (-1)(3\alpha+8) + 1(\alpha^2-6\alpha-16)$$

Combining like terms, we obtain:

$$= \alpha + 3\alpha + 8 + \alpha^2 - 6\alpha - 16$$

Simplifying further leads to

$$= \alpha^2 - 2\alpha - 8 = 0$$

Factoring this quadratic expression yields

$$(\alpha - 4)(\alpha + 2) = 0$$

Therefore, $$\alpha = 4$$ or $$\alpha = -2$$, and summing these values gives $$4 + (-2) = 2$$.

Hence, the correct answer is 2.

The given vectors are:

$$\vec{a} = \lambda\hat{i} + \mu\hat{j} + 4\hat{k}$$

$$\vec{b} = -2\hat{i} + 4\hat{j} - 2\hat{k}$$

$$\vec{c} = 2\hat{i} + 3\hat{j} + \hat{k}$$

Since the vectors $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$ are coplanar, their scalar triple product must be zero:

$$\begin{vmatrix} \lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1 \end{vmatrix} = 0$$

Expanding the determinant along the first row:

$$\lambda(4(1) - (-2)(3)) - \mu((-2)(1) - (-2)(2)) + 4((-2)(3) - 4(2)) = 0$$

$$\lambda(4 + 6) - \mu(-2 + 4) + 4(-6 - 8) = 0$$

$$10\lambda - 2\mu - 56 = 0 \implies 5\lambda - \mu = 28$$

Rearranging for $$\mu$$ gives:

$$\mu = 5\lambda - 28$$

Next, the magnitude of the projection of $$\vec{a}$$ on $$\vec{b}$$ is given as $$\sqrt{54}$$ units:

$$\frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} = \sqrt{54}$$

Computing the dot product $$\vec{a} \cdot \vec{b}$$ and the magnitude $$|\vec{b}|$$:

$$\vec{a} \cdot \vec{b} = (\lambda)(-2) + (\mu)(4) + (4)(-2) = -2\lambda + 4\mu - 8$$

$$|\vec{b}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24}$$

Substitute these values into the projection equation:

$$\frac{|-2\lambda + 4\mu - 8|}{\sqrt{24}} = \sqrt{54}$$

$$|-2\lambda + 4\mu - 8| = \sqrt{54 \times 24} = \sqrt{1296} = 36$$

Dividing by 2 inside the absolute value yields:

$$|-\lambda + 2\mu - 4| = 18$$

This gives two possible cases to explore:

Case 1: $$-\lambda + 2\mu - 4 = 18 \implies -\lambda + 2\mu = 22$$

Substituting $$\mu = 5\lambda - 28$$ into this linear relation:

$$-\lambda + 2(5\lambda - 28) = 22 \implies 9\lambda = 78 \implies \lambda = \frac{26}{3}$$

$$\mu = 5\left(\frac{26}{3}\right) - 28 = \frac{46}{3}$$

Thus, the first possible value for the sum is:

$$(\lambda + \mu)_1 = \frac{26}{3} + \frac{46}{3} = \frac{72}{3} = 24$$

Case 2: $$-\lambda + 2\mu - 4 = -18 \implies -\lambda + 2\mu = -14$$

Substituting $$\mu = 5\lambda - 28$$ into this linear relation:

$$-\lambda + 2(5\lambda - 28) = -14 \implies 9\lambda = 42 \implies \lambda = \frac{14}{3}$$

$$\mu = 5\left(\frac{14}{3}\right) - 28 = -\frac{14}{3}$$

Thus, the second possible value for the sum is:

$$(\lambda + \mu)_2 = \frac{14}{3} + \left(-\frac{14}{3}\right) = 0$$

Summing all possible values of $$\lambda + \mu$$:

$$\text{Sum} = 24 + 0 = 24$$

Conclusion:

The sum of all possible values of $$\lambda + \mu$$ is equal to 24.

Vectors are coplanar if their scalar triple product (determinant) is zero:

$$\begin{vmatrix} 1 & 1 & a \\ 1 & b & 1 \\ c & 1 & 1 \end{vmatrix} = 0$$

Expanding along the first row:

$$1(b - 1) - 1(1 - c) + a(1 - bc) = 0$$

$$b - 1 - 1 + c + a - abc = 0 \implies \mathbf{a + b + c - 2 = abc}$$

$$\begin{vmatrix} a+b & c & c \\ a & b+c & a \\ b & b & c+a \end{vmatrix} = 0$$

Using row operations ($$R_1 \to R_1 - R_2 - R_3$$):

$$\begin{vmatrix} 0 & -2b & -2a \\ a & b+c & a \\ b & b & c+a \end{vmatrix} = 0$$

Expanding along the first row:

$$2b(a(c+a) - ab) - 2a(ab - b(b+c)) = 0$$

$$2b(ac + a^2 - ab) - 2a(ab - b^2 - bc) = 0$$

Divide by $$2b$$ (assuming $$b \neq 0$$):

$$(ac + a^2 - ab) - \frac{a}{b}(ab - b^2 - bc) = 0$$

$$ac + a^2 - ab - a^2 + ab + ac = 0 \implies \mathbf{2ac = 0}$$

This implies $$a=0$$ or $$c=0$$.

If $$a=0$$, substituting into our first equation ($$a + b + c - 2 = abc$$):

$$0 + b + c - 2 = 0 \implies \mathbf{b + c = 2}$$

The expression we need is $$6(a + b + c)$$:

$$6(0 + 2) = \mathbf{12}$$

Given $$\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$$, and the condition $$|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$$ with $$\vec{b} \cdot \vec{c} = 0$$.

From $$|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$$, squaring both sides:

$$|\vec{a} + \vec{b}|^2 + 2(\vec{a} + \vec{b}) \cdot \vec{c} + |\vec{c}|^2 = |\vec{a} + \vec{b}|^2 - 2(\vec{a} + \vec{b}) \cdot \vec{c} + |\vec{c}|^2$$

This gives $$4(\vec{a} + \vec{b}) \cdot \vec{c} = 0$$, so $$(\vec{a} + \vec{b}) \cdot \vec{c} = 0$$.

Since $$\vec{b} \cdot \vec{c} = 0$$, we get $$\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} = 0$$, which gives $$\vec{a} \cdot \vec{c} = 0$$.

Statement A: $$|\vec{a} + \lambda\vec{c}| \geq |\vec{a}|$$ for all $$\lambda \in \mathbb{R}$$.

$$|\vec{a} + \lambda\vec{c}|^2 = |\vec{a}|^2 + 2\lambda(\vec{a} \cdot \vec{c}) + \lambda^2|\vec{c}|^2 = |\vec{a}|^2 + \lambda^2|\vec{c}|^2$$

Since $$\vec{a} \cdot \vec{c} = 0$$ and $$\lambda^2|\vec{c}|^2 \geq 0$$, we have $$|\vec{a} + \lambda\vec{c}|^2 \geq |\vec{a}|^2$$.

Therefore $$|\vec{a} + \lambda\vec{c}| \geq |\vec{a}|$$. Statement A is correct.

Statement B: $$\vec{a}$$ and $$\vec{c}$$ are always parallel.

Since $$\vec{a} \cdot \vec{c} = 0$$ and $$\vec{c} \neq \vec{0}$$, $$\vec{a}$$ and $$\vec{c}$$ are perpendicular, not parallel. Statement B is false.

The answer is Option C: only (A) is correct.

We are given $$\vec{\alpha}=4\hat{i}+3\hat{j}+5\hat{k}$$ and $$\vec{\beta}=1\hat{i}+2\hat{j}-4\hat{k}$$, and we decompose $$\vec{\beta}$$ into a component parallel to $$\vec{\alpha}$$, denoted by $$\vec{\beta_1}$$, and a component perpendicular to $$\vec{\alpha}$$, denoted by $$\vec{\beta_2}$$, so that $$\vec{\beta}=\vec{\beta_1}+\vec{\beta_2}$$. Our goal is to evaluate $$5\vec{\beta_2}\cdot(\hat{i}+\hat{j}+\hat{k})$$.

By the definition of vector projection, we have

$$\vec{\beta_1}=\frac{\vec{\beta}\cdot\vec{\alpha}}{|\vec{\alpha}|^2}\,\vec{\alpha}$$

We first compute the scalar product

$$\vec{\beta}\cdot\vec{\alpha}=(1)(4)+(2)(3)+(-4)(5)=4+6-20=-10$$

and the squared magnitude of $$\vec{\alpha}$$

$$|\vec{\alpha}|^2=4^2+3^2+5^2=16+9+25=50$$

Substituting these values into the formula for $$\vec{\beta_1}$$ gives

$$\vec{\beta_1}=\frac{-10}{50}\,\vec{\alpha}=-\frac{1}{5}(4\hat{i}+3\hat{j}+5\hat{k})=-\frac{4}{5}\hat{i}-\frac{3}{5}\hat{j}-\hat{k}$$

The perpendicular component then follows from the relation $$\vec{\beta_2}=\vec{\beta}-\vec{\beta_1}$$, which yields

$$\vec{\beta_2}=\left(1+\frac{4}{5}\right)\hat{i}+\left(2+\frac{3}{5}\right)\hat{j}+(-4+1)\hat{k}=\frac{9}{5}\hat{i}+\frac{13}{5}\hat{j}-3\hat{k}$$

Multiplying $$\vec{\beta_2}$$ by 5 gives

$$5\vec{\beta_2}=9\hat{i}+13\hat{j}-15\hat{k}$$

Finally, taking the dot product with $$\hat{i}+\hat{j}+\hat{k}$$ leads to

$$5\vec{\beta_2}\cdot(\hat{i}+\hat{j}+\hat{k})=9+13-15=7$$

Hence the required value is 7.

We need to find $$|\vec{a} \cdot \vec{b}|$$ given the conditions $$\vec{a} \cdot \vec{c} = 7$$, $$2(\vec{b} \cdot \vec{c}) + 43 = 0$$, and $$\vec{a} \times \vec{c} = \vec{b} \times \vec{c}$$.

$$\vec{a} \times \vec{c} = \vec{b} \times \vec{c} \implies (\vec{a} - \vec{b}) \times \vec{c} = \vec{0}$$

This means $$\vec{a} - \vec{b}$$ is parallel to $$\vec{c}$$, so $$\vec{c} = t(\vec{a} - \vec{b})$$ for some scalar $$t$$.

$$\vec{a} - \vec{b} = (\hat{i} + 2\hat{j} + \lambda\hat{k}) - (3\hat{i} - 5\hat{j} - \lambda\hat{k}) = -2\hat{i} + 7\hat{j} + 2\lambda\hat{k}$$

From $$\vec{a} \cdot \vec{c} = 7$$:

$$t[(-2)(1) + (7)(2) + (2\lambda)(\lambda)] = 7 \implies t(12 + 2\lambda^2) = 7 \quad \text{...(i)}$$

From $$2(\vec{b} \cdot \vec{c}) + 43 = 0 \implies \vec{b} \cdot \vec{c} = -\frac{43}{2}$$:

$$t[(-2)(3) + (7)(-5) + (2\lambda)(-\lambda)] = -\frac{43}{2} \implies t(-41 - 2\lambda^2) = -\frac{43}{2} \quad \text{...(ii)}$$

Dividing (i) by (ii):

$$\frac{12 + 2\lambda^2}{-(41 + 2\lambda^2)} = \frac{7}{-43/2} = -\frac{14}{43}$$

$$43(12 + 2\lambda^2) = 14(41 + 2\lambda^2)$$

$$516 + 86\lambda^2 = 574 + 28\lambda^2$$

$$58\lambda^2 = 58 \implies \lambda^2 = 1$$

$$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-5) + (\lambda)(-\lambda) = 3 - 10 - \lambda^2 = 3 - 10 - 1 = -8$$

$$|\vec{a} \cdot \vec{b}| = 8$$

The correct answer is $$8$$.

Given $$\vec{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$$, $$\vec{b} = \hat{i} + \hat{k}$$, and $$\vec{c} = \hat{i} + 2\hat{j} - 3\hat{k}$$. We need to find $$|\vec{r}|$$ where $$\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$$ and $$\vec{r} \cdot \vec{b} = 0$$.

From the condition $$\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$$ we have $$\vec{r} \times \vec{a} - \vec{c} \times \vec{a} = \vec{0}$$, which implies $$(\vec{r} - \vec{c}) \times \vec{a} = \vec{0}$$. Hence $$\vec{r} - \vec{c}$$ is parallel to $$\vec{a}$$, and we can write $$\vec{r} = \vec{c} + \lambda \vec{a}$$ for some scalar $$\lambda$$.

Substituting the vectors gives $$\vec{r} = (1 + 2\lambda)\hat{i} + (2 - 7\lambda)\hat{j} + (-3 + 5\lambda)\hat{k}$$.

Applying the dot product condition $$\vec{r} \cdot \vec{b} = 0$$ yields $$(1 + 2\lambda)\cdot 1 + (2 - 7\lambda)\cdot 0 + (-3 + 5\lambda)\cdot 1 = 0$$, so $$(1 + 2\lambda) + (-3 + 5\lambda) = 0$$ which leads to $$-2 + 7\lambda = 0$$ and hence $$\lambda = \frac{2}{7}$$.

Substituting $$\lambda = \frac{2}{7}$$ gives $$\vec{r} = \left(1 + \frac{4}{7}\right)\hat{i} + (2 - 2)\hat{j} + \left(-3 + \frac{10}{7}\right)\hat{k} = \frac{11}{7}\hat{i} + 0\hat{j} - \frac{11}{7}\hat{k}$$.

The magnitude of $$\vec{r}$$ is $$|\vec{r}| = \sqrt{\left(\frac{11}{7}\right)^2 + 0 + \left(\frac{11}{7}\right)^2} = \frac{11}{7}\sqrt{2}$$, so $$|\vec{r}| = \frac{11\sqrt{2}}{7} \approx 2.22$$. Therefore, $$|\vec{r}| = \frac{11\sqrt{2}}{7}$$.

We are given $$\vec{b} \cdot \vec{c} = 0$$ and $$\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$$. Since the BAC-CAB identity states that $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$$, substituting yields the equation relating these expressions for the triple product.

Therefore, we have

$$\vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b}) = \frac{\vec{b} - \vec{c}}{2}\;. $$

Comparing coefficients of $$\vec{b}$$ and $$\vec{c}$$ from this equation gives $$\vec{a} \cdot \vec{c} = \frac{1}{2}$$ and $$\vec{a} \cdot \vec{b} = \frac{1}{2}$$. Next, we note that $$\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b} = \frac{1}{2}$$ and, from above, $$\vec{b} \cdot \vec{c} = 0$$.

Applying the scalar quadruple product identity,

$$ (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})\;, $$

and substituting the known values leads to

$$ (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) - (\vec{a} \cdot \vec{d})(0) = \frac{1}{4}\;. $$

Therefore, the correct answer is Option D: $$\frac{1}{4}$$.

Applying the Lagrange identity for vectors, we have $$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$$. Substituting the given magnitudes yields $$ (\sqrt{48})^2 + (\vec{a} \cdot \vec{b})^2 = (\sqrt{14})^2 \times (\sqrt{6})^2$$. Simplifying these expressions gives $$48 + (\vec{a} \cdot \vec{b})^2 = 14 \times 6 = 84$$ and therefore $$ (\vec{a} \cdot \vec{b})^2 = 84 - 48 = 36\,. $$

Let the position vectors of the points $$A$$, $$B$$, $$C$$, and $$D$$ be denoted as $$\vec{OA}$$, $$\vec{OB}$$, $$\vec{OC}$$, and $$\vec{OD}$$ respectively. We first find the vectors $$\vec{AB}$$, $$\vec{AC}$$, and $$\vec{AD}$$:

$$\vec{AB} = \vec{OB} - \vec{OA} = (\lambda - 1)\vec{a} - 2\vec{b} + 3\vec{c}$$

$$\vec{AC} = \vec{OC} - \vec{OA} = -2\vec{a} + 3\vec{b} - 4\vec{c}$$

$$\vec{AD} = \vec{OD} - \vec{OA} = \vec{a} - 3\vec{b} + 5\vec{c}$$

Since the vectors $$\vec{AB}$$, $$\vec{AC}$$, and $$\vec{AD}$$ are coplanar and the base vectors $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$ are non-coplanar, the scalar triple product $$[\vec{AB} \quad \vec{AC} \quad \vec{AD}]$$ must equal zero. This is equivalent to setting the determinant of their coefficients to zero:

$$\begin{vmatrix} \lambda - 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} = 0$$

Expanding the determinant along the first row:

$$(\lambda - 1) \begin{vmatrix} 3 & -4 \\ -3 & 5 \end{vmatrix} - (-2) \begin{vmatrix} -2 & -4 \\ 1 & 5 \end{vmatrix} + 3 \begin{vmatrix} -2 & 3 \\ 1 & -3 \end{vmatrix} = 0$$

Evaluating each minor:

- $$\begin{vmatrix} 3 & -4 \\ -3 & 5 \end{vmatrix} = (3)(5) - (-4)(-3) = 15 - 12 = 3$$

- $$\begin{vmatrix} -2 & -4 \\ 1 & 5 \end{vmatrix} = (-2)(5) - (-4)(1) = -10 + 4 = -6$$

- $$\begin{vmatrix} -2 & 3 \\ 1 & -3 \end{vmatrix} = (-2)(-3) - (3)(1) = 6 - 3 = 3$$

Substituting these values back into the expanded equation:

$$3(\lambda - 1) + 2(-6) + 3(3) = 0$$

$$3\lambda - 3 - 12 + 9 = 0$$

$$3\lambda - 6 = 0$$

$$3\lambda = 6 \implies \lambda = 2$$

Conclusion:

The value of $$\lambda$$ is equal to 2.

We are given the vector $$\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$$, which is rotated through a right angle in a plane that passes through the y-axis to yield $$\vec{b}$$, and we wish to find the projection of $$3\vec{a} + \sqrt{2}\vec{b}$$ onto $$\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$$.

Since the rotation passes through the y-axis ($$\hat{j}$$ direction), it occurs in the plane spanned by $$\vec{a}$$ and $$\hat{j}$$, we write $$\vec{b} = \lambda\vec{a} + \mu\hat{j}$$.

Enforcing perpendicularity between $$\vec{a}$$ and $$\vec{b}$$ gives $$\vec{a} \cdot \vec{b} = \lambda|\vec{a}|^2 + \mu(\vec{a} \cdot \hat{j}) = 0$$. Since $$|\vec{a}|^2 = (-1)^2 + 2^2 + 1^2 = 6$$ and $$\vec{a} \cdot \hat{j} = 2$$, it follows that $$6\lambda + 2\mu = 0 \implies \mu = -3\lambda$$.

Substituting into $$\vec{b}$$ yields $$\vec{b} = \lambda(-1, 2, 1) + (-3\lambda)(0, 1, 0) = (-\lambda, 2\lambda - 3\lambda, \lambda) = (-\lambda, -\lambda, \lambda)$$. Imposing $$|\vec{b}| = |\vec{a}|$$ leads to $$|\vec{b}|^2 = \lambda^2 + \lambda^2 + \lambda^2 = 3\lambda^2 = 6 \implies \lambda = \pm\sqrt{2}$$.

To select the correct sign, note that during the $$90°$$ rotation from $$\vec{a} = (-1, 2, 1)$$ the path must pass through the $$+\hat{j}$$ direction. For $$\lambda = -\sqrt{2}$$ one finds $$\vec{b} = (\sqrt{2}, \sqrt{2}, -\sqrt{2})$$ and $$\sqrt{2}\vec{b} = (2, 2, -2)$$, whereas for $$\lambda = \sqrt{2}$$ one finds $$\vec{b} = (-\sqrt{2}, -\sqrt{2}, \sqrt{2})$$ and $$\sqrt{2}\vec{b} = (-2, -2, 2)$$. Only the choice $$\lambda = -\sqrt{2}$$ gives a positive $$y$$-component in $$\vec{b}$$, so we take $$\lambda = -\sqrt{2}$$.

Therefore, it follows that $$3\vec{a} = 3(-1, 2, 1) = (-3, 6, 3)$$ and, with this choice of $$\lambda$$, $$\sqrt{2}\vec{b} = (2, 2, -2)$$, so $$3\vec{a} + \sqrt{2}\vec{b} = (-3 + 2, 6 + 2, 3 - 2) = (-1, 8, 1)$$. Since $$|\vec{c}| = \sqrt{25 + 16 + 9} = 5\sqrt{2}$$, the projection of $$3\vec{a} + \sqrt{2}\vec{b}$$ on $$\vec{c}$$ is $$\frac{(3\vec{a} + \sqrt{2}\vec{b}) \cdot \vec{c}}{|\vec{c}|} = \frac{(-1)(5) + (8)(4) + (1)(3)}{5\sqrt{2}} = \frac{-5 + 32 + 3}{5\sqrt{2}} = \frac{30}{5\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$$.

Therefore, the projection is $$\mathbf{3\sqrt{2}}$$.

We have vectors $$\vec{a} = 6\hat{i} + 9\hat{j} + 12\hat{k}$$, $$\vec{b} = \alpha\hat{i} + 11\hat{j} - 2\hat{k}$$, and the condition $$\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$$.

Since $$\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$$ implies $$\vec{a} \times (\vec{c} - \vec{b}) = \vec{0}$$, the vector $$\vec{c} - \vec{b}$$ must be parallel to $$\vec{a}$$. Thus we write $$\vec{c} = \vec{b} + \lambda\vec{a}$$ for some scalar $$\lambda$$, which in components gives $$\vec{c} = (\alpha + 6\lambda)\hat{i} + (11 + 9\lambda)\hat{j} + (-2 + 12\lambda)\hat{k}$$.

Applying the condition $$\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5$$ leads to $$(\alpha + 6\lambda) - 2(11 + 9\lambda) + (-2 + 12\lambda) = 5$$, which simplifies to $$\alpha + 6\lambda - 22 - 18\lambda - 2 + 12\lambda = 5$$ and hence to $$\alpha - 24 = 5\implies \alpha = 29$$.

Next, using $$\vec{a} \cdot \vec{c} = -12$$ and substituting $$\alpha = 29$$ into the expression for $$\vec{c}$$ gives $$6(29 + 6\lambda) + 9(11 + 9\lambda) + 12(-2 + 12\lambda) = -12$$. This expands to $$174 + 36\lambda + 99 + 81\lambda - 24 + 144\lambda = -12$$, so $$249 + 261\lambda = -12$$, yielding $$\lambda = -1$$.

Substituting $$\alpha = 29$$ and $$\lambda = -1$$ into $$\vec{c} = (\alpha + 6\lambda)\hat{i} + (11 + 9\lambda)\hat{j} + (-2 + 12\lambda)\hat{k}$$ gives $$\vec{c} = 23\hat{i} + 2\hat{j} - 14\hat{k}$$.

Finally, the required dot product is $$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 + (-14) = 11$$, so the answer is $$11$$.

Given: $$\vec{v} = \alpha\hat{i} + 2\hat{j} - 3\hat{k}$$, $$\vec{w} = 2\alpha\hat{i} + \hat{j} - \hat{k}$$, $$|\vec{u}| = \alpha > 0$$.

The scalar triple product $$[\vec{u}\ \vec{v}\ \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$$.

First, compute $$\vec{v} \times \vec{w}$$.

Next, find minimum of $$[\vec{u}\ \vec{v}\ \vec{w}]$$.

The minimum value of $$\vec{u} \cdot (\vec{v} \times \vec{w})$$ with $$|\vec{u}| = \alpha$$ is $$-\alpha|\vec{v} \times \vec{w}|$$, occurring when $$\vec{u}$$ is antiparallel to $$\vec{v} \times \vec{w}$$.

Given: minimum = $$-\alpha\sqrt{3401}$$

Now, find $$|\vec{u} \cdot \hat{i}|^2$$.

When the minimum occurs, $$\vec{u} = -\frac{\alpha}{|\vec{v} \times \vec{w}|}(\vec{v} \times \vec{w}) = -\frac{10}{\sqrt{3401}}(1, -50, -30)$$

Since $$\gcd(100, 3401) = 1$$ (as $$3401 = 34 \times 100 + 1$$), we have $$m = 100$$, $$n = 3401$$.

Therefore, $$m + n = 3501$$.

The area of a quadrilateral $$ABCD$$ in 3D space, defined by its vertices, can be found using the magnitude of the cross product of its diagonal vectors $$\vec{AC}$$ and $$\vec{BD}$$:

$$\text{Area} = \frac{1}{2} |\vec{AC} \times \vec{BD}|$$

$$\vec{AC} = \vec{C} - \vec{A} = (2 - 2)\hat{i} + (3 - 6)\hat{j} + (-1 - 2)\hat{k}$$

$$\vec{AC} = 0\hat{i} - 3\hat{j} - 3\hat{k}$$

$$\vec{BD} = \vec{D} - \vec{B} = (4 - (-4))\hat{i} + (5 - 0)\hat{j} + (0 - \lambda)\hat{k}$$

$$\vec{BD} = 8\hat{i} + 5\hat{j} - \lambda\hat{k}$$

$$\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda \end{vmatrix}$$

$$\vec{AC} \times \vec{BD} = \hat{i}((-3)(-\lambda) - (-3)(5)) - \hat{j}(0(-\lambda) - (-3)(8)) + \hat{k}(0(5) - (-3)(8))$$

$$\vec{AC} \times \vec{BD} = (3\lambda + 15)\hat{i} - 24\hat{j} + 24\hat{k}$$

$$\frac{1}{2} |\vec{AC} \times \vec{BD}| = 18$$

$$|\vec{AC} \times \vec{BD}| = 36$$

$$\sqrt{(3\lambda + 15)^2 + (-24)^2 + (24)^2} = 36$$

$$(3\lambda + 15)^2 + 576 + 576 = 1296$$

$$(3\lambda + 15)^2 = 144$$

$$3\lambda + 15 = \pm 12$$

$$3\lambda = -15 \pm 12$$

$$\implies \lambda = -1, -9$$ 

$$\lambda = -1$$ as per the given condition

$$5 - 6(-1) = 5 + 6 = 11$$

We are given

$$\vec a = 3\hat i+\hat j-\hat k,\qquad \vec b = \hat i+b_2\hat j+b_3\hat k,\qquad \vec c = c_1\hat i+c_2\hat j+c_3\hat k,\qquad b_2b_3\gt 0.$$ The conditions are

1. $$\vec a\cdot\vec b = 0,$$

2. $$\begin{pmatrix} 0 & -c_3 & c_2 \\ c_3 & 0 & -c_1 \\ -c_2 & c_1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{pmatrix} 3-c_1 \\ 1-c_2 \\ -1-c_3 \end{pmatrix}.$$ All assertions have to be checked one by one.

Step 1: Using $$\vec a\cdot\vec b=0$$
$$\vec a\cdot\vec b = 3(1)+1(b_2)+(-1)(b_3)=0 \;\Longrightarrow\; 3+b_2-b_3=0 \;\Longrightarrow\; b_3 = 3+b_2.$$

The extra restriction $$b_2b_3\gt 0$$ now reads

Either $$b_2\gt 0\quad(\text{then }b_3=3+b_2\gt 3),$$
or $$b_2\lt -3\quad(\text{then }b_3=3+b_2\lt 0).$$

Step 2: Translating the matrix equation into three scalar equations

The given matrix equation is equivalent to

$$\begin{cases} -b_2c_3 + (3+b_2)c_2 + c_1 = 3 & -(A)\\[2pt] \;c_3 - (3+b_2)c_1 + c_2 = 1 & -(B)\\[2pt] \;c_3 + b_2c_1 - c_2 = -1 & -(C) \end{cases}$$

Step 3: Solving (B) and (C)

Adding (B) and (C): $$2c_3 - 3c_1 = 0 \;\Longrightarrow\; c_3=\dfrac32\,c_1.$$
Subtracting (C) from (B): $$2c_2-(3+2b_2)c_1=2 \;\Longrightarrow\; c_2 = 1+\dfrac{3+2b_2}{2}\,c_1.$$ Put $$k=\dfrac{3+2b_2}{2}\quad\bigl(\text{so }c_2=1+k\,c_1\bigr).$$

Step 4: Using equation (A) to determine $$c_1$$

Insert the relations for $$c_2$$ and $$c_3$$ in (A):

$$-b_2\!\left(\tfrac32c_1\right) +(3+b_2)(1+k\,c_1)+c_1 = 3.$$

Simplifying, $$\bigl(b_2^2+3b_2+\tfrac{11}{2}\bigr)c_1 = -b_2.$$ Define $$D=b_2^2+3b_2+\tfrac{11}{2}>0,$$ then

$$c_1 = -\dfrac{\,b_2\,}{D},\qquad c_2 = 1+k\,c_1,\qquad c_3 = \dfrac32c_1.$$

Step 5: Dot-product $$\vec b\cdot\vec c$$

$$\vec b\cdot\vec c = 1\cdot c_1 + b_2c_2 + (3+b_2)c_3.$$ Substitute $$c_2=1+k\,c_1,\; c_3=\tfrac32c_1$$ to get

$$\vec b\cdot\vec c = b_2 + \bigl(1+b_2k+\tfrac32(3+b_2)\bigr)c_1.$$ But the coefficient $$1+b_2k+\tfrac32(3+b_2)=D$$ is exactly the same $$D$$ introduced above. Hence $$\vec b\cdot\vec c = b_2 - D\,\dfrac{b_2}{D}=0.$$ So statement B ( $$\vec b\cdot\vec c=0$$ ) is TRUE.

Step 6: Magnitude of $$\vec b$$

$$|\vec b|^2 = 1+b_2^2+b_3^2 = 1+b_2^2+(3+b_2)^2 = 2b_2^2+6b_2+10 = 2b_2(b_2+3)+10.$$ Because $$b_2$$ and $$(b_2+3)$$ have the same sign (see Step 1), the product $$2b_2(b_2+3)$$ is \emph{strictly positive}. Therefore

$$|\vec b|^2 \gt 10 \;\Longrightarrow\; |\vec b| \gt \sqrt{10}.$$ Hence statement C is TRUE.

Step 7: Magnitude of $$\vec c$$

Using the expressions found for $$c_1,c_2,c_3$$:

$$\begin{aligned} |\vec c|^2 &= c_1^2+c_2^2+c_3^2 \\[2pt] &= c_1^2+\bigl(1+k\,c_1\bigr)^2+\bigl(\tfrac32c_1\bigr)^2 \\[2pt] &= 1-\dfrac{2k\,b_2}{D}+\dfrac{(3.25+k^2)b_2^2}{D^2}\\[2pt] &= \dfrac{11/2}{\,D\,}. \end{aligned}$$ (Every algebraic term collapses to the simple fraction shown.)

Because $$D=b_2^2+3b_2+\tfrac{11}{2}\ge \tfrac{13}{4} \;(\text{always})$$ we get

$$|\vec c|^2 \le \dfrac{11}{2}\;\Big/\;\dfrac{13}{4} = \dfrac{22}{13} \lt 11.$$ Thus $$|\vec c|\le \sqrt{11},$$ so statement D is TRUE.

Step 8: Checking $$\vec a\cdot\vec c$$

$$\vec a\cdot\vec c = 3c_1+c_2-c_3 = 1+\bigl(3+b_2\bigr)c_1 = 1-\dfrac{b_2(3+b_2)}{D} = \dfrac{11/2}{\,D\,},$$ the same positive quantity obtained in Step 7. Hence $$\vec a\cdot\vec c\neq 0,$$ so statement A is FALSE.

Conclusion
The correct statements are
Option B ( $$\vec b\cdot\vec c=0$$ ),
Option C ( $$|\vec b|\gt\sqrt{10}$$ ) and
Option D ( $$|\vec c|\le\sqrt{11}$$ ).

Answer: Option A is false; Options B, C and D are TRUE.

Given $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$ and $$\vec{b} = \hat{i} + \hat{j} - \hat{k}$$.

The first vector is $$(\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k}$$. Since $$\vec{a} \cdot \hat{i} = 1$$, $$\vec{a} \cdot \hat{j} = 1$$, $$\vec{a} \cdot \hat{k} = 1$$, this gives $$\hat{i} + \hat{j} + \hat{k} = \vec{a}$$.

The second vector is $$(\vec{b} \cdot \hat{i})\hat{i} + (\vec{b} \cdot \hat{j})\hat{j} + (\vec{b} \cdot \hat{k})\hat{k}$$. Since $$\vec{b} \cdot \hat{i} = 1$$, $$\vec{b} \cdot \hat{j} = 1$$, $$\vec{b} \cdot \hat{k} = -1$$, this gives $$\hat{i} + \hat{j} - \hat{k} = \vec{b}$$.

The third vector is $$\vec{c} = \hat{i} + \hat{j} - 2\hat{k}$$.

So we have three vectors: $$\vec{a} = (1, 1, 1)$$, $$\vec{b} = (1, 1, -1)$$, $$\vec{c} = (1, 1, -2)$$.

To check coplanarity, we compute the scalar triple product $$[\vec{a}\;\vec{b}\;\vec{c}]$$:

$$[\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & 1 & -2 \end{vmatrix}$$

Expanding along the first row: $$= 1(1 \cdot (-2) - (-1) \cdot 1) - 1(1 \cdot (-2) - (-1) \cdot 1) + 1(1 \cdot 1 - 1 \cdot 1)$$

$$= 1(-2 + 1) - 1(-2 + 1) + 1(0) = (-1) - (-1) + 0 = 0$$

Since the scalar triple product is zero, the three vectors are coplanar.

The correct answer is Option B: the vectors are coplanar.

We are given four vectors $$\vec{u} = \langle 2, 3, 9 \rangle$$, $$\vec{v} = \langle 5, 2, 1 \rangle$$, $$\vec{w} = \langle 1, \lambda, 8 \rangle$$, and $$\vec{z} = \langle \lambda, 2, 3 \rangle$$ that are coplanar. Since these are position vectors (or free vectors) that are coplanar, one of them can be expressed as a linear combination of the others. Equivalently, since four vectors in $$\mathbb{R}^3$$ are always linearly dependent, we interpret coplanarity as the condition that the vectors $$\vec{v} - \vec{u}$$, $$\vec{w} - \vec{u}$$, and $$\vec{z} - \vec{u}$$ are coplanar, meaning their scalar triple product is zero.

We compute $$\vec{v} - \vec{u} = \langle 3, -1, -8 \rangle$$, $$\vec{w} - \vec{u} = \langle -1, \lambda - 3, -1 \rangle$$, and $$\vec{z} - \vec{u} = \langle \lambda - 2, -1, -6 \rangle$$.

The scalar triple product equals the determinant: $$\begin{vmatrix} 3 & -1 & -8 \\ -1 & \lambda - 3 & -1 \\ \lambda - 2 & -1 & -6 \end{vmatrix} = 0$$.

Expanding along the first row: $$3[(\lambda - 3)(-6) - (-1)(-1)] - (-1)[(-1)(-6) - (-1)(\lambda - 2)] + (-8)[(-1)(-1) - (\lambda - 3)(\lambda - 2)] = 0$$.

We simplify each term. The first term gives $$3[-6\lambda + 18 - 1] = 3(-6\lambda + 17) = -18\lambda + 51$$. The second term gives $$1[6 - (-\lambda + 2)] = 1[6 + \lambda - 2] = \lambda + 4$$. The third term gives $$-8[1 - (\lambda^2 - 5\lambda + 6)] = -8[1 - \lambda^2 + 5\lambda - 6] = -8[-\lambda^2 + 5\lambda - 5] = 8\lambda^2 - 40\lambda + 40$$.

Adding all terms: $$-18\lambda + 51 + \lambda + 4 + 8\lambda^2 - 40\lambda + 40 = 0$$, which gives $$8\lambda^2 - 57\lambda + 95 = 0$$.

Using the quadratic formula: $$\lambda = \frac{57 \pm \sqrt{3249 - 3040}}{16} = \frac{57 \pm \sqrt{209}}{16}$$. However, we need the product of the roots. By Vieta's formulas, the product of all possible values of $$\lambda$$ is $$\frac{95}{8}$$.

Hence, the correct answer is Option D.

We have the vectors $$\vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}$$, $$\vec{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$$ and $$\vec{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$$. Computing $$\vec{AB} = \vec{b} - \vec{a} = \hat{i} + (\alpha-4)\hat{j} + \hat{k}$$ and $$\vec{AC} = \vec{c} - \vec{a} = 2\hat{i} - 6\hat{j} + 2\hat{k}$$, we set $$\vec{AB} = \lambda\vec{AC}$$. From $$1 = 2\lambda$$ we get $$\lambda = 1/2$$, and $$\alpha - 4 = -6\times\frac12 = -3$$ gives $$\alpha = 1$$. Verifying $$1 = 2\times\frac12 = 1$$, so the points are collinear when $$\alpha = 1$$.

Since $$\alpha = 1$$ makes them collinear, the smallest positive integer for non-collinearity is $$\alpha = 2$$.

With $$\alpha = 2$$, the points are $$B = (2,2,4)$$ and $$C = (3,-2,5)$$. The midpoint of $$BC$$ is $$M = \left(\frac{2+3}{2},\frac{2-2}{2},\frac{4+5}{2}\right) = \left(\frac{5}{2},0,\frac{9}{2}\right)$$.

Writing $$A = (1,4,3)$$, the length of the median $$AM$$ is $$AM = \sqrt{\left(\frac{5}{2}-1\right)^2 + (0-4)^2 + \left(\frac{9}{2}-3\right)^2} = \sqrt{\left(\frac{3}{2}\right)^2 + 16 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \sqrt{\frac{9 + 64 + 9}{4}} = \frac{\sqrt{82}}{2}\,. $$

Therefore, the answer is Option A: $$\frac{\sqrt{82}}{2}$$.

Given,

$$\vec{BC}=\vec a,\qquad \vec{CA}=\vec b,\qquad \vec{AB}=\vec c$$

Also,

$$|\vec a|=6\sqrt2,\qquad |\vec b|=2\sqrt3,\qquad \vec b\cdot\vec c=12$$

Since

$$\vec a+\vec b+\vec c=0,$$

we get

$$\vec c=-(\vec a+\vec b)$$

Now,

$$\vec b\cdot\vec c=12$$

$$\vec b\cdot(-\vec a-\vec b)=12$$

$$-\vec a\cdot\vec b-|\vec b|^2=12$$

$$-\vec a\cdot\vec b-(2\sqrt3)^2=12$$

$$-\vec a\cdot\vec b-12=12$$

$$\vec a\cdot\vec b=-24$$

Now find

$$|\vec c|^2$$

$$|\vec c|^2=|\vec a+\vec b|^2$$

$$=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b$$

$$=(6\sqrt2)^2+(2\sqrt3)^2+2(-24)$$

$$=72+12-48$$

$$=36$$

Hence,

$$|\vec c|=6$$

Now check statement $$S_1.$$

Since

$$\vec b+\vec c=-\vec a,$$

$$\vec a\times(\vec b+\vec c)=\vec a\times(-\vec a)=0$$

Therefore,

$$|\vec a\times(\vec b+\vec c)|=0$$

Hence,

$$|\vec a\times(\vec b+\vec c)|\times|\vec b-\vec c|=0$$

which is not equal to

$$6(2\sqrt2-1)$$

Therefore,

$$S_1$$ is false.

Now check $$S_2.$$

Angle $$ABC$$ is the angle between

$$\vec{BA}=-\vec c$$

and

$$\vec{BC}=\vec a$$

Hence,

$$\cos\angle ABC=\frac{(-\vec c)\cdot\vec a}{|\vec c||\vec a|}$$

Now,

$$\vec c=-(\vec a+\vec b)$$

$$\vec a\cdot\vec c=\vec a\cdot(-\vec a-\vec b)$$

$$=-|\vec a|^2-\vec a\cdot\vec b$$

$$=-72-(-24)$$

$$=-48$$

Thus,

$$(-\vec c)\cdot\vec a=48$$

Therefore,

$$\cos\angle ABC=\frac{48}{6\cdot6\sqrt2}$$

$$=\frac{48}{36\sqrt2}$$

$$=\frac4{3\sqrt2}$$

$$=\frac{2\sqrt2}{3}$$

$$=\sqrt{\frac89}$$

which is not equal to

$$\sqrt{\frac23}$$

Hence,

$$S_2$$ is false.

Therefore,

$$\boxed{\text{Both }S_1\text{ and }S_2\text{ are false}}$$.

Let the angle between the unit vectors $$\hat a$$ and $$\hat b$$ be $$\theta$$.

Given,

$$\left|(\hat a+\hat b)+2(\hat a\times\hat b)\right|=2$$

Now,

$$\hat a+\hat b$$ lies in the plane containing $$\hat a,\hat b$$ while

$$\hat a\times\hat b$$ is perpendicular to that plane.

Hence,

$$ (\hat a+\hat b)\perp(\hat a\times\hat b) $$

Therefore, using

$$|\vec u+\vec v|^2=|\vec u|^2+|\vec v|^2$$

for perpendicular vectors,

$$\left|(\hat a+\hat b)+2(\hat a\times\hat b)\right|^2=|\hat a+\hat b|^2+|2(\hat a\times\hat b)|^2$$

Given modulus is $$2,$$ so

$$|\hat a+\hat b|^2+|2(\hat a\times\hat b)|^2=4$$

Now,

$$|\hat a+\hat b|^2=|\hat a|^2+|\hat b|^2+2\hat a\cdot\hat b$$

Since $$|\hat a|=|\hat b|=1,$$

$$|\hat a+\hat b|^2=1+1+2\cos\theta$$

Also,

$$|\hat a\times\hat b|=|\hat a||\hat b|\sin\theta=\sin\theta$$

Hence,

$$|2(\hat a\times\hat b)|^2=4\sin^2\theta$$

Substituting,

$$2+2\cos\theta+4\sin^2\theta=4$$

Using

$$\sin^2\theta=1-\cos^2\theta,$$

we get

$$2+2\cos\theta+4(1-\cos^2\theta)=4$$

$$2+2\cos\theta+4-4\cos^2\theta=4$$

$$4\cos^2\theta-2\cos\theta-2=0$$

Dividing by $$2,$$

$$2\cos^2\theta-\cos\theta-1=0$$

Factorising,

$$(2\cos\theta+1)(\cos\theta-1)=0$$

Thus,

$$\cos\theta=1\qquad \text{or}\qquad \cos\theta=-\frac12$$

Since the vectors are not parallel in the given condition, we take

$$\cos\theta=-\frac12$$

Hence,

$$\theta=\frac{2\pi}{3}=120^\circ$$

Now verify Statement $$(S1)$$:

$$2|\hat a\times\hat b|=|\hat a-\hat b|$$

LHS:

$$2|\hat a\times\hat b|=2\sin120^\circ$$

$$=2\cdot\frac{\sqrt3}{2}=\sqrt3$$

RHS:

$$|\hat a-\hat b|=\sqrt{|\hat a|^2+|\hat b|^2-2\hat a\cdot\hat b}$$

$$=\sqrt{1+1-2\cos120^\circ}$$

$$=\sqrt{2-2\left(-\frac12\right)}$$

$$=\sqrt3$$

Thus,

$$2|\hat a\times\hat b|=|\hat a-\hat b|$$

Hence, $$(S1)$$ is true.

Now verify Statement $$(S2)$$:

The projection of $$\hat a$$ on $$\hat a+\hat b$$ is

$$\frac{\hat a\cdot(\hat a+\hat b)}{|\hat a+\hat b|}$$

$$=\frac{\hat a\cdot\hat a+\hat a\cdot\hat b}{\sqrt{|\hat a|^2+|\hat b|^2+2\hat a\cdot\hat b}}$$

$$=\frac{1+\cos120^\circ}{\sqrt{2+2\cos120^\circ}}$$

$$=\frac{1-\frac12}{\sqrt{2-1}}$$

$$=\frac12$$

Hence, $$(S2)$$ is also true.

Therefore, both statements are true.

Hence, the correct answer is $$\boxed{\text{Both (S1) and (S2) are true}}$$.

We need the angle between $$\vec{u} = a(\log_e b)\hat{i} - 6\hat{j} + 3\hat{k}$$ and $$\vec{v} = (\log_e b)\hat{i} + 2\hat{j} + 2a(\log_e b)\hat{k}$$ to be acute, where $$b > 1$$.

For the angle to be acute, we need $$\vec{u} \cdot \vec{v} > 0$$. Computing the dot product:

$$\vec{u} \cdot \vec{v} = a(\log_e b)^2 - 12 + 6a(\log_e b)$$

Let $$t = \log_e b$$. Since $$b > 1$$, we have $$t > 0$$. So we need:

$$at^2 + 6at - 12 > 0$$

for ALL $$t > 0$$ (since the condition must hold for every valid $$b > 1$$, meaning every $$t > 0$$).

Let $$f(t) = at^2 + 6at - 12$$. We need $$f(t) > 0$$ for all $$t > 0$$.

Case 1: If $$a = 0$$, then $$f(t) = -12 < 0$$. Not valid.

Case 2: If $$a > 0$$, the parabola opens upward. At $$t = 0$$: $$f(0) = -12 < 0$$. Since the function is continuous and negative at $$t = 0$$, by continuity $$f(t) < 0$$ for small positive $$t$$. So $$f(t) > 0$$ for ALL $$t > 0$$ fails.

Case 3: If $$a < 0$$, the parabola opens downward. As $$t \to \infty$$, $$f(t) \to -\infty$$. So $$f(t) > 0$$ for all $$t > 0$$ is impossible here too (since eventually $$f$$ becomes negative).

So there is no value of $$a$$ for which the angle is acute for all $$b > 1$$.

Hence, $$S = \emptyset$$, and the correct answer is Option B.

We need to find all values of $$t$$ for which the vectors $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ are linearly independent (i.e., $$\alpha\vec{a} + \beta\vec{b} + \gamma\vec{c} = \vec{0}$$ implies $$\alpha = \beta = \gamma = 0$$).

Since the vectors are linearly independent if and only if the determinant of the matrix formed by their components is non-zero, we consider:

$$D = \begin{vmatrix} 1+t & 1-t & 1 \\ 1-t & 1+t & 2 \\ t & -t & 1 \end{vmatrix}$$

Expanding this determinant along the first row gives:

$$D = (1+t)[(1+t)(1) - 2(-t)] - (1-t)[(1-t)(1) - 2t] + 1[(-t)(1-t) - t(1+t)]$$

This simplifies to:

$$= (1+t)(1+t+2t) - (1-t)(1-t-2t) + (-t+t^2 - t - t^2)$$

$$= (1+t)(1+3t) - (1-t)(1-3t) + (-2t)$$

$$= (1 + 3t + t + 3t^2) - (1 - 3t - t + 3t^2) - 2t$$

$$= (1 + 4t + 3t^2) - (1 - 4t + 3t^2) - 2t$$

$$= 8t - 2t = 6t$$

From the above, the determinant is $$D = 6t$$, so the vectors are linearly independent exactly when $$D \neq 0$$, that is, when $$6t \neq 0$$, or equivalently $$t \neq 0$$.

Therefore, the set of all such values of $$t$$ is $$\mathbb{R} - \{0\}$$.

The answer is Option C.

We have $$\vec{a} = 3\hat{i} + \hat{j}$$, $$\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$$, and the relation $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda\vec{c}$$.

We apply the BAC-CAB identity for the vector triple product: $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$$.

We compute $$\vec{a} \cdot \vec{b} = 3(1) + 1(2) + 0(1) = 5$$. So the equation becomes:

$$\vec{b}(\vec{a} \cdot \vec{c}) - 5\vec{c} = \vec{b} + \lambda\vec{c}$$

Rearranging: $$(\vec{a} \cdot \vec{c} - 1)\vec{b} = (5 + \lambda)\vec{c}$$.

Since $$\vec{b}$$ and $$\vec{c}$$ are non-parallel, neither can be expressed as a scalar multiple of the other. The only way the equation $$(\vec{a} \cdot \vec{c} - 1)\vec{b} = (5 + \lambda)\vec{c}$$ can hold with $$\vec{b}$$ and $$\vec{c}$$ non-parallel is if both coefficients are zero. That is, $$\vec{a} \cdot \vec{c} - 1 = 0$$ and $$5 + \lambda = 0$$.

From the second condition, $$\lambda = -5$$.

Hence, the correct answer is Option A.

We need to find $$\alpha$$ given that the projection of $$\vec{a} \times \vec{b}$$ on $$-\hat{i} + 2\hat{j} - 2\hat{k}$$ is 30.

First, with $$\vec{a} = \alpha\hat{i} + \hat{j} - \hat{k}$$ and $$\vec{b} = 2\hat{i} + \hat{j} - \alpha\hat{k}$$, we compute $$\vec{a} \times \vec{b}$$:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & -1 \\ 2 & 1 & -\alpha \end{vmatrix}$$

$$= \hat{i}(1 \cdot (-\alpha) - (-1) \cdot 1) - \hat{j}(\alpha \cdot (-\alpha) - (-1) \cdot 2) + \hat{k}(\alpha \cdot 1 - 1 \cdot 2)$$

$$= \hat{i}(-\alpha + 1) - \hat{j}(-\alpha^2 + 2) + \hat{k}(\alpha - 2)$$

$$= (1 - \alpha)\hat{i} + (\alpha^2 - 2)\hat{j} + (\alpha - 2)\hat{k}$$

Next, let $$\vec{c} = -\hat{i} + 2\hat{j} - 2\hat{k}$$ so that $$|\vec{c}| = \sqrt{1 + 4 + 4} = 3$$, and compute the projection of $$\vec{a} \times \vec{b}$$ on $$\vec{c}$$:

Projection = $$\dfrac{(\vec{a} \times \vec{b}) \cdot \vec{c}}{|\vec{c}|}$$

$$ (\vec{a} \times \vec{b}) \cdot \vec{c} = (1 - \alpha)(-1) + (\alpha^2 - 2)(2) + (\alpha - 2)(-2)$$

$$= -(1 - \alpha) + 2\alpha^2 - 4 - 2\alpha + 4$$

$$= -1 + \alpha + 2\alpha^2 - 4 - 2\alpha + 4$$

$$= 2\alpha^2 - \alpha - 1$$

Then substituting into the projection formula gives $$\dfrac{2\alpha^2 - \alpha - 1}{3} = 30$$, which leads to $$2\alpha^2 - \alpha - 1 = 90$$ and hence $$2\alpha^2 - \alpha - 91 = 0$$.

Finally, solving $$2\alpha^2 - \alpha - 91 = 0$$ yields $$\alpha = \dfrac{1 \pm \sqrt{1 + 728}}{4} = \dfrac{1 \pm \sqrt{729}}{4} = \dfrac{1 \pm 27}{4}$$, so $$\alpha = \dfrac{28}{4} = 7$$ or $$\alpha = \dfrac{-26}{4}$$ (rejected since $$\alpha > 0$$).

The correct answer is Option D: $$\alpha = 7$$.

Let the diagonals of the parallelogram be

$$\vec a \quad \text{and} \quad \vec b$$

Area of a parallelogram in terms of diagonals is

$$\text{Area}=\frac12|\vec a\times \vec b|$$

Given,

$$\frac12|\vec a\times \vec b|=2\sqrt2$$

Hence,

$$|\vec a\times \vec b|=4\sqrt2$$

Now,

$$|\vec a|=1$$

and

$$|\vec a\cdot \vec b|=|\vec a\times \vec b|$$

Let the angle between $$\vec a$$ and $$\vec b$$ be $$\theta$$.

Then,

$$|\vec a||\vec b|\cos\theta=|\vec a||\vec b|\sin\theta$$

$$\tan\theta=1$$

Since the angle is acute,

$$\theta=\frac{\pi}{4}$$

Also,

$$|\vec a\times \vec b| = |\vec a||\vec b|\sin\frac{\pi}{4} $$

$$4\sqrt2=1\cdot |\vec b|\cdot \frac1{\sqrt2}$$

$$|\vec b|=8$$

Given,

$$\vec c=2\sqrt2(\vec a\times \vec b)-2\vec b$$

Now,

$$\vec b\cdot(\vec a\times \vec b)=0$$

Therefore,

$$\vec b\cdot\vec c = -2|\vec b|^2 = -2(64) = -128 $$

Also,

$$|2\sqrt2(\vec a\times \vec b)|=16$$

and

$$|-2\vec b|=16$$

Since these two vectors are perpendicular,

$$|\vec c| = \sqrt{16^2+16^2} = 16\sqrt2$$

Let the angle between $$\vec b$$ and $$\vec c$$ be $$\alpha$$.

Then,

$$\cos\alpha = \frac{\vec b\cdot\vec c}{|\vec b||\vec c|} = \frac{-128}{8\cdot16\sqrt2} = -\frac1{\sqrt2} $$

Hence,

$$\alpha=\frac{3\pi}{4}$$

Therefore, the correct answer is

$$\boxed{\text{Option D }\frac{3\pi}{4}}$$

We are given $$\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$$, $$\vec{a} \times \vec{b} = 2\hat{i} - \hat{k}$$, and $$\vec{a} \cdot \vec{b} = 3$$. To determine $$\vec{b}$$, let $$\vec{b}=(b_1,b_2,b_3)$$. The cross product in component form becomes

Equating components gives

$$-b_3 - 2b_2 = 2\quad(i),\qquad 2b_1 - b_3 = 0\quad(ii),\qquad b_1 + b_2 = -1\quad(iii).$$

From (ii) we have $$b_3 = 2b_1$$, and from (iii) $$b_2 = -1 - b_1$$. Substituting these into the dot-product condition $$\vec{a}\cdot\vec{b}=3$$ yields

$$b_1 - b_2 + 2b_3 = 3 \;\implies\; b_1 + 1 + b_1 + 4b_1 = 3 \;\implies\; 6b_1 = 2 \;\implies\; b_1 = \tfrac{1}{3}.$$

Thus

$$\vec{b}=\Bigl(\tfrac{1}{3},\,-\tfrac{4}{3},\,\tfrac{2}{3}\Bigr).$$

Subtracting this from $$\vec{a}$$ gives

$$\vec{a}-\vec{b}=\Bigl(\tfrac{2}{3},\;\tfrac{1}{3},\;\tfrac{4}{3}\Bigr).$$

Next, the scalar product of $$\vec{b}$$ with $$\vec{a}-\vec{b}$$ is

The magnitude of $$\vec{a}-\vec{b}$$ is

Therefore, the projection of $$\vec{b}$$ onto $$\vec{a}-\vec{b}$$ is

The answer is Option A: $$\dfrac{2}{\sqrt{21}}$$.

We are given $$\vec{a} = \hat{i} + \hat{j} - \hat{k}$$ and $$\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}$$. We need to find the number of vectors $$\vec{b}$$ such that $$\vec{b} \times \vec{c} = \vec{a}$$ and $$|\vec{b}| \in \{1, 2, \ldots, 10\}$$.

First, for the cross product equation $$\vec{b} \times \vec{c} = \vec{a}$$ to have a solution, $$\vec{a}$$ must be perpendicular to $$\vec{c}$$ because $$\vec{b} \times \vec{c}$$ is always perpendicular to $$\vec{c}$$.

Now, computing the dot product gives $$\vec{a} \cdot \vec{c} = (1)(2) + (1)(-3) + (-1)(2) = 2 - 3 - 2 = -3 \neq 0$$.

Since $$\vec{a} \cdot \vec{c} \neq 0$$, $$\vec{a}$$ is not perpendicular to $$\vec{c}$$, and consequently there is no vector $$\vec{b}$$ satisfying $$\vec{b} \times \vec{c} = \vec{a}$$.

The number of such vectors $$\vec{b}$$ is $$0$$.

The answer is Option A: $$0$$.

We are given $$|\vec{a}| = 9$$ and for every $$(x, y) \neq (0, 0)$$, the vector $$(x\vec{a} + y\vec{b})$$ is perpendicular to $$(6y\vec{a} - 18x\vec{b})$$.

Since perpendicular vectors have zero dot product, we have $$(x\vec{a} + y\vec{b}) \cdot (6y\vec{a} - 18x\vec{b}) = 0$$. Expanding the dot product gives $$6xy|\vec{a}|^2 - 18x^2(\vec{a} \cdot \vec{b}) + 6y^2(\vec{a} \cdot \vec{b}) - 18xy|\vec{b}|^2 = 0$$, which can be grouped as $$6xy(|\vec{a}|^2 - 3|\vec{b}|^2) + 6(y^2 - 3x^2)(\vec{a} \cdot \vec{b}) = 0$$.

Because this identity holds for all $$(x, y) \neq (0,0)$$, the coefficients of the monomials must each vanish. Substituting $$x = 1, y = 0$$ into the grouped expression yields $$6(0 - 3)(\vec{a} \cdot \vec{b}) = 0$$, so $$\vec{a} \cdot \vec{b} = 0$$. Similarly, substituting $$x = 1, y = 1$$ gives $$6(|\vec{a}|^2 - 3|\vec{b}|^2) = 0$$, hence $$|\vec{a}|^2 = 3|\vec{b}|^2$$.

Using the given value of $$|\vec{a}|$$, we have $$81 = 3|\vec{b}|^2$$, which implies $$|\vec{b}|^2 = 27$$ and therefore $$|\vec{b}| = 3\sqrt{3}$$.

Since $$\vec{a} \cdot \vec{b} = 0$$, the vectors are perpendicular (θ = 90°). It follows that $$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(90°) = 9 \times 3\sqrt{3} \times 1 = 27\sqrt{3}$$.

The answer is Option B.

We have two unit vectors $$\hat{a}$$ and $$\hat{b}$$ with the angle between them equal to $$\frac{\pi}{4}$$, so $$|\hat{a}| = |\hat{b}| = 1$$ and $$\hat{a} \cdot \hat{b} = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$. Also, $$|\hat{a} \times \hat{b}| = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$.

We define $$\vec{p} = \hat{a} + \hat{b}$$ and $$\vec{q} = \hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})$$, and we need to find $$164\cos^2\theta$$ where $$\theta$$ is the angle between $$\vec{p}$$ and $$\vec{q}$$.

We compute $$|\vec{p}|^2 = |\hat{a}|^2 + 2(\hat{a}\cdot\hat{b}) + |\hat{b}|^2 = 1 + \frac{2}{\sqrt{2}} + 1 = 2 + \sqrt{2}$$.

Now we find $$|\vec{q}|^2$$. We note that $$\hat{a} \times \hat{b}$$ is perpendicular to both $$\hat{a}$$ and $$\hat{b}$$, so $$\hat{a} \cdot (\hat{a} \times \hat{b}) = 0$$ and $$\hat{b} \cdot (\hat{a} \times \hat{b}) = 0$$.

$$|\vec{q}|^2 = |\hat{a} + 2\hat{b}|^2 + 4|\hat{a}\times\hat{b}|^2 = \left(1 + 4\cdot\frac{1}{\sqrt{2}} + 4\right) + 4\cdot\frac{1}{2} = 5 + 2\sqrt{2} + 2 = 7 + 2\sqrt{2} + 2\sqrt{2} = 5 + 4\cdot\frac{1}{\sqrt{2}} + 4 + 2$$

Let me compute this more carefully. $$|\hat{a}+2\hat{b}|^2 = 1 + 4(\hat{a}\cdot\hat{b}) + 4 = 5 + \frac{4}{\sqrt{2}} = 5 + 2\sqrt{2}$$. And $$|2(\hat{a}\times\hat{b})|^2 = 4\cdot\frac{1}{2} = 2$$. Since $$(\hat{a}+2\hat{b})$$ is perpendicular to $$2(\hat{a}\times\hat{b})$$, we get $$|\vec{q}|^2 = 5 + 2\sqrt{2} + 2 = 7 + 2\sqrt{2}$$.

Next, $$\vec{p}\cdot\vec{q} = (\hat{a}+\hat{b})\cdot(\hat{a}+2\hat{b}+2(\hat{a}\times\hat{b})) = (\hat{a}+\hat{b})\cdot(\hat{a}+2\hat{b}) + 0$$, since $$(\hat{a}+\hat{b})\cdot(\hat{a}\times\hat{b}) = 0$$ (as $$\hat{a}\times\hat{b}$$ is perpendicular to both $$\hat{a}$$ and $$\hat{b}$$).

$$\vec{p}\cdot\vec{q} = \hat{a}\cdot\hat{a} + 2(\hat{a}\cdot\hat{b}) + \hat{b}\cdot\hat{a} + 2(\hat{b}\cdot\hat{b}) = 1 + \frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 2 = 3 + \frac{3}{\sqrt{2}} = 3 + \frac{3\sqrt{2}}{2}$$

Now $$\cos^2\theta = \frac{(\vec{p}\cdot\vec{q})^2}{|\vec{p}|^2\,|\vec{q}|^2}$$.

We compute the numerator: $$\left(3 + \frac{3\sqrt{2}}{2}\right)^2 = 9 + 2\cdot 3\cdot\frac{3\sqrt{2}}{2} + \frac{9\cdot 2}{4} = 9 + 9\sqrt{2} + \frac{9}{2} = \frac{27}{2} + 9\sqrt{2} = \frac{27 + 18\sqrt{2}}{2}$$.

The denominator: $$(2+\sqrt{2})(7+2\sqrt{2}) = 14 + 4\sqrt{2} + 7\sqrt{2} + 2\cdot 2 = 14 + 11\sqrt{2} + 4 = 18 + 11\sqrt{2}$$.

So $$\cos^2\theta = \frac{27 + 18\sqrt{2}}{2(18+11\sqrt{2})}$$.

We rationalize by multiplying numerator and denominator by $$(18 - 11\sqrt{2})$$:

Denominator factor: $$(18+11\sqrt{2})(18-11\sqrt{2}) = 324 - 242 = 82$$.

Numerator factor: $$(27+18\sqrt{2})(18-11\sqrt{2}) = 486 - 297\sqrt{2} + 324\sqrt{2} - 198\cdot 2 = 486 - 396 + (324-297)\sqrt{2} = 90 + 27\sqrt{2}$$.

So $$\cos^2\theta = \frac{90+27\sqrt{2}}{2\cdot 82} = \frac{90+27\sqrt{2}}{164}$$.

Therefore $$164\cos^2\theta = 90 + 27\sqrt{2}$$.

Hence, the correct answer is Option A.

Let $$\hat{a}$$ and $$\hat{b}$$ be unit vectors, and $$\vec{c}$$ be a vector such that the angle between $$\hat{a}$$ and $$\vec{c}$$ is $$\frac{\pi}{12}$$, and $$\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})$$.

Take the magnitude squared of both sides.

$$|\hat{b}|^2 = |\vec{c} + 2(\vec{c} \times \hat{a})|^2$$

$$1 = |\vec{c}|^2 + 4|\vec{c} \times \hat{a}|^2 + 4\vec{c} \cdot (\vec{c} \times \hat{a})$$

Simplify the cross terms.

The scalar triple product $$\vec{c} \cdot (\vec{c} \times \hat{a}) = 0$$ (since it involves two identical vectors in the triple product).

$$|\vec{c} \times \hat{a}|^2 = |\vec{c}|^2 \sin^2\left(\frac{\pi}{12}\right)$$

Substitute and simplify.

$$1 = |\vec{c}|^2 + 4|\vec{c}|^2 \sin^2\left(\frac{\pi}{12}\right) = |\vec{c}|^2\left(1 + 4\sin^2\frac{\pi}{12}\right)$$

Compute $$\sin^2\left(\frac{\pi}{12}\right)$$.

Using the identity $$\sin^2\theta = \frac{1 - \cos 2\theta}{2}$$:

$$\sin^2\frac{\pi}{12} = \frac{1 - \cos\frac{\pi}{6}}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}$$

Find the coefficient.

$$1 + 4 \cdot \frac{2 - \sqrt{3}}{4} = 1 + 2 - \sqrt{3} = 3 - \sqrt{3}$$

$$|\vec{c}|^2 = \frac{1}{3 - \sqrt{3}}$$

Compute $$|6\vec{c}|^2$$.

$$|6\vec{c}|^2 = 36|\vec{c}|^2 = \frac{36}{3 - \sqrt{3}}$$

Rationalizing: $$= \frac{36(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})} = \frac{36(3 + \sqrt{3})}{9 - 3} = \frac{36(3 + \sqrt{3})}{6} = 6(3 + \sqrt{3})$$

Answer: Option B

We begin by noting that $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$ with each $$a_i > 0$$ and that it makes equal angles with the coordinate axes. Consequently, $$a_1 = a_2 = a_3 = k$$ for some $$k > 0$$. The projection of $$\vec{a}$$ on $$3\hat{i} + 4\hat{j}$$ is given by
$$ \frac{\vec{a} \cdot (3\hat{i} + 4\hat{j})}{|3\hat{i} + 4\hat{j}|} = \frac{3k + 4k}{5} = \frac{7k}{5} = 7 $$
which implies $$k = 5$$ and hence $$\vec{a} = 5\hat{i} + 5\hat{j} + 5\hat{k}$$.

Next, to determine $$\vec{b}$$ we observe that rotating $$\vec{a}$$ by $$90°$$ preserves its magnitude, so $$|\vec{b}| = |\vec{a}| = 5\sqrt{3}$$ while also ensuring $$\vec{a} \cdot \vec{b} = 0$$. Because $$\vec{a}$$, $$\vec{b}$$, and the x-axis are coplanar, $$\vec{b}$$ lies in the plane spanned by $$\vec{a}$$ and $$\hat{i}$$. Thus we may write
$$ \vec{b} = \alpha\vec{a} + \beta\hat{i} $$

Then, imposing the orthogonality condition $$\vec{a} \cdot \vec{b} = 0$$ yields
$$ \alpha|\vec{a}|^2 + \beta(\vec{a} \cdot \hat{i}) = 0 \implies 75\alpha + 5\beta = 0 \implies \beta = -15\alpha $$
Hence $$\vec{b} = \alpha(5,5,5) + (-15\alpha)(1,0,0) = \alpha(-10,5,5)$$. Next, enforcing the magnitude condition $$|\vec{b}| = 5\sqrt{3}$$ gives
$$ \alpha^2(100+25+25) = 75 \implies 150\alpha^2 = 75 \implies \alpha = \pm\frac{1}{\sqrt{2}} $$

Therefore
$$ \vec{b} = \pm\frac{1}{\sqrt{2}}(-10,5,5) $$
and its projection on $$3\hat{i} + 4\hat{j}$$ is
$$ \text{Projection} = \frac{\vec{b} \cdot (3\hat{i} + 4\hat{j})}{5} = \frac{\pm\frac{1}{\sqrt{2}}(-30+20)}{5} = \frac{\pm\frac{1}{\sqrt{2}}\cdot(-10)}{5} = \frac{\mp10}{5\sqrt{2}} = \mp\sqrt{2} $$
Taking the positive value yields $$\sqrt{2}$$.

The answer is Option B: $$\sqrt{2}$$.

We have $$\vec{a} = \alpha\hat{i} + 2\hat{j} - \hat{k}$$ and $$\vec{b} = -2\hat{i} + \alpha\hat{j} + \hat{k}$$.

Compute $$\vec{a} \times \vec{b}$$:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -1 \\ -2 & \alpha & 1 \end{vmatrix}$$

$$= \hat{i}(2 + \alpha) - \hat{j}(\alpha - 2) + \hat{k}(\alpha^2 + 4)$$

Find the area condition:

$$|\vec{a} \times \vec{b}|^2 = (\alpha + 2)^2 + (\alpha - 2)^2 + (\alpha^2 + 4)^2$$

$$= \alpha^2 + 4\alpha + 4 + \alpha^2 - 4\alpha + 4 + \alpha^4 + 8\alpha^2 + 16$$

$$= \alpha^4 + 10\alpha^2 + 24$$

Given area = $$\sqrt{15(\alpha^2 + 4)}$$, so:

$$|\vec{a} \times \vec{b}|^2 = 15(\alpha^2 + 4) = 15\alpha^2 + 60$$

Setting equal: $$\alpha^4 + 10\alpha^2 + 24 = 15\alpha^2 + 60$$

$$\alpha^4 - 5\alpha^2 - 36 = 0$$

Solve for $$\alpha$$:

Let $$u = \alpha^2$$: $$u^2 - 5u - 36 = 0$$

$$(u - 9)(u + 4) = 0$$

Since $$u = \alpha^2 \geq 0$$: $$\alpha^2 = 9$$, so $$\alpha = \pm 3$$.

Compute the required expression:

$$|\vec{a}|^2 = \alpha^2 + 4 + 1 = 9 + 5 = 14$$

$$|\vec{b}|^2 = 4 + \alpha^2 + 1 = 4 + 9 + 1 = 14$$

$$\vec{a} \cdot \vec{b} = -2\alpha + 2\alpha - 1 = -1$$

$$2|\vec{a}|^2 + (\vec{a} \cdot \vec{b})|\vec{b}|^2 = 2(14) + (-1)(14) = 28 - 14 = 14$$

The correct answer is Option D: $$\boxed{14}$$.

Given $$\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$$, $$\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$$, and $$\vec{c} = \hat{i} - \hat{j} + \hat{k}$$.

First, we express $$\vec{v}$$ in the plane of $$\vec{a}$$ and $$\vec{b}$$ as $$\vec{v} = \lambda\vec{a} + \mu\vec{b} = (\lambda + 2\mu)\hat{i} + (\lambda - 3\mu)\hat{j} + (2\lambda + \mu)\hat{k}$$.

Next, using the condition $$\vec{v} \cdot \hat{j} = 7$$ yields $$\lambda - 3\mu = 7 \quad \cdots (1)$$.

Since the projection of $$\vec{v}$$ on $$\vec{c}$$ is $$\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{2}{\sqrt{3}},$$ we note that $$|\vec{c}| = \sqrt{1+1+1} = \sqrt{3}$$ and

$$\vec{v} \cdot \vec{c} = (\lambda + 2\mu)(1) + (\lambda - 3\mu)(-1) + (2\lambda + \mu)(1) = \lambda + 2\mu - \lambda + 3\mu + 2\lambda + \mu = 2\lambda + 6\mu.$$

Therefore, $$\frac{2\lambda + 6\mu}{\sqrt{3}} = \frac{2}{\sqrt{3}} \implies 2\lambda + 6\mu = 2 \implies \lambda + 3\mu = 1 \quad \cdots (2).$$

Now, from (1) we have $$\lambda = 7 + 3\mu$$, and substituting into (2) gives $$7 + 3\mu + 3\mu = 1 \implies 6\mu = -6 \implies \mu = -1$$ and hence $$\lambda = 7 + 3(-1) = 4$$.

Substituting these into the expression for $$\vec{v}$$ yields $$\vec{v} = (4 + 2(-1))\hat{i} + (4 - 3(-1))\hat{j} + (8 + (-1))\hat{k} = 2\hat{i} + 7\hat{j} + 7\hat{k}$$.

Finally, computing the required dot product gives $$\vec{v} \cdot (\hat{i} + \hat{k}) = 2 + 7 = 9$$.

Hence the correct answer is Option D: $$9$$.

We have three coplanar concurrent vectors $$\vec{a}, \vec{b}, \vec{c}$$ such that the angle between any two of them is the same. Since they lie in a plane and radiate from a common point, the only configuration where all pairwise angles are equal is when each pair makes an angle of $$120°$$ with every other pair (as $$\frac{360°}{3} = 120°$$).

Let $$|\vec{a}| = a$$, $$|\vec{b}| = b$$, and $$|\vec{c}| = c$$. We are given that the product of magnitudes satisfies $$abc = 14$$.

Since all three vectors are coplanar, every cross product among them points in the same direction — perpendicular to the common plane. Let $$\hat{n}$$ be the unit normal to this plane. With the angle between each pair being $$120°$$ and $$\sin 120° = \frac{\sqrt{3}}{2}$$, we have $$\vec{a} \times \vec{b} = \frac{\sqrt{3}}{2}\,ab\,\hat{n}$$, $$\vec{b} \times \vec{c} = \frac{\sqrt{3}}{2}\,bc\,\hat{n}$$, and $$\vec{c} \times \vec{a} = \frac{\sqrt{3}}{2}\,ca\,\hat{n}$$. (All cross products point along $$\hat{n}$$ or $$-\hat{n}$$ depending on orientation, but the dot products of parallel vectors use the same sign consistently.)

Now we compute each dot product in the given expression. We have $$(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2}\,ab \cdot \frac{\sqrt{3}}{2}\,bc = \frac{3}{4}\,ab^2c$$. Similarly, $$(\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = \frac{3}{4}\,abc^2$$, and $$(\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = \frac{3}{4}\,a^2bc$$.

Adding all three terms gives $$\frac{3}{4}\,abc(a + b + c) = \frac{3}{4} \times 14 \times (a + b + c) = \frac{21}{2}(a + b + c)$$.

Setting this equal to 168, we obtain $$\frac{21}{2}(a + b + c) = 168$$, which yields $$a + b + c = \frac{168 \times 2}{21} = 16$$.

Hence, the correct answer is Option C.

We need to find the obtuse angle between $$\vec{a}$$ (parallel to the line of intersection of two planes) and $$\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$$.

The first plane is determined by vectors $$\hat{i}$$ and $$\hat{i} + \hat{j}$$.

Normal to the first plane: $$\vec{n_1} = \hat{i} \times (\hat{i} + \hat{j}) = \hat{i} \times \hat{i} + \hat{i} \times \hat{j} = \vec{0} + \hat{k} = \hat{k}$$

So the first plane has normal $$\hat{k}$$, i.e., it is the $$xy$$-plane ($$z = 0$$).

The second plane is determined by vectors $$\hat{i} - \hat{j}$$ and $$\hat{i} + \hat{k}$$.

Normal to the second plane:

$$\vec{n_2} = (\hat{i} - \hat{j}) \times (\hat{i} + \hat{k}) = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 0 & 1\end{vmatrix}$$

$$= \hat{i}(-1-0) - \hat{j}(1-0) + \hat{k}(0+1) = -\hat{i} - \hat{j} + \hat{k}$$

$$\vec{a}$$ is parallel to $$\vec{n_1} \times \vec{n_2}$$:

$$\vec{n_1} \times \vec{n_2} = \hat{k} \times (-\hat{i} - \hat{j} + \hat{k}) = -(\hat{k} \times \hat{i}) - (\hat{k} \times \hat{j}) + (\hat{k} \times \hat{k})$$

$$= -\hat{j} - (-\hat{i}) + \vec{0} = \hat{i} - \hat{j}$$

So $$\vec{a}$$ is parallel to $$\hat{i} - \hat{j}$$.

$$\cos\alpha = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{(1)(1) + (-1)(-2) + (0)(2)}{\sqrt{1+1}\cdot\sqrt{1+4+4}} = \dfrac{1+2}{(\sqrt{2})(3)} = \dfrac{3}{3\sqrt{2}} = \dfrac{1}{\sqrt{2}}$$

The acute angle is $$\alpha = \dfrac{\pi}{4}$$.

The obtuse angle is $$\pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$$.

The correct answer is Option A: $$\dfrac{3\pi}{4}$$.

Given planes,

$$2x+ky-5z=1$$

and

$$3kx-ky+z=5$$

These two planes are mutually perpendicular.

Hence, product of normals is zero.

Normal vectors are

$$\vec n_1=(2,k,-5)$$

and

$$\vec n_2=(3k,-k,1)$$

Therefore,

$$\vec n_1\cdot\vec n_2=0$$

$$2(3k)+k(-k)+(-5)(1)=0$$

$$6k-k^2-5=0$$

$$k^2-6k+5=0$$

$$(k-1)(k-5)=0$$

Given

$$k<3,$$

hence

$$k=1$$

So the planes become

$$2x+y-5z=1$$

and

$$3x-y+z=5$$

Any plane through their line of intersection is

$$2x+y-5z-1+\lambda(3x-y+z-5)=0$$

$$ (2+3\lambda)x+(1-\lambda)y+(-5+\lambda)z-(1+5\lambda)=0 $$

The plane intercepts unit length on positive $$x$$-axis.

Hence, putting

$$y=0,\qquad z=0,\qquad x=1$$

must satisfy the plane.

Therefore,

$$2+3\lambda-(1+5\lambda)=0$$

$$1-2\lambda=0$$

$$\lambda=\frac12$$

Substituting,

$$\frac72x+\frac12y-\frac92z-\frac72=0$$

Multiplying by $$2,$$

$$7x+y-9z-7=0$$

Now find the $$y$$-intercept.

Putting

$$x=0,\qquad z=0,$$

$$y-7=0$$

$$y=7$$

Hence, the intercept on the $$y$$-axis is

$$\boxed{7}$$.

Let $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$. We are given two conditions.

Use the perpendicularity condition:

$$\vec{a} \cdot (3\hat{i} + \frac{1}{2}\hat{j} + 2\hat{k}) = 0$$

$$3a_1 + \frac{a_2}{2} + 2a_3 = 0 \quad \cdots (1)$$

Use the cross product condition:

$$\vec{a} \times (2\hat{i} + \hat{k}) = 2\hat{i} - 13\hat{j} - 4\hat{k}$$

$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 2 & 0 & 1 \end{vmatrix} = (a_2 - 0)\hat{i} - (a_1 - 2a_3)\hat{j} + (0 - 2a_2)\hat{k}$$

$$= a_2\hat{i} - (a_1 - 2a_3)\hat{j} - 2a_2\hat{k}$$

Comparing components:

$$a_2 = 2 \quad \cdots (2)$$

$$-(a_1 - 2a_3) = -13 \implies a_1 - 2a_3 = 13 \quad \cdots (3)$$

$$-2a_2 = -4 \implies a_2 = 2 \quad \checkmark$$

Solve the system:

From equation (1): $$3a_1 + 1 + 2a_3 = 0 \implies 3a_1 + 2a_3 = -1 \quad \cdots (4)$$

From equation (3): $$a_1 = 13 + 2a_3$$

Substituting into (4): $$3(13 + 2a_3) + 2a_3 = -1$$

$$39 + 6a_3 + 2a_3 = -1$$

$$8a_3 = -40 \implies a_3 = -5$$

$$a_1 = 13 + 2(-5) = 3$$

So $$\vec{a} = 3\hat{i} + 2\hat{j} - 5\hat{k}$$.

Find the projection:

Projection of $$\vec{a}$$ on $$\vec{c} = 2\hat{i} + 2\hat{j} + \hat{k}$$:

$$\text{proj} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \frac{6 + 4 - 5}{\sqrt{4 + 4 + 1}} = \frac{5}{3}$$

The correct answer is Option C: $$\boxed{\dfrac{5}{3}}$$.

Given $$|\vec{a}| = 4$$, $$|\vec{b}| = 3$$ and $$\theta \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$$, we wish to determine the value of $$|(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b})|^2 + 4|\vec{a} \cdot \vec{b}|^2$$.

Expanding the cross product, we have $$(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}.$$ Since $$\vec{a} \times \vec{a} = \vec{0}$$, $$\vec{b} \times \vec{b} = \vec{0}$$, and $$\vec{b} \times \vec{a} = -\,\vec{a} \times \vec{b},$$ this reduces to $$2(\vec{a} \times \vec{b}).$$

Therefore, the squared magnitude of this cross product is $$|2(\vec{a} \times \vec{b})|^2 = 4\,|\vec{a} \times \vec{b}|^2 = 4\,|\vec{a}|^2\,|\vec{b}|^2\sin^2\theta.$$

Similarly, the second term is $$4\,|\vec{a} \cdot \vec{b}|^2 = 4\,|\vec{a}|^2\,|\vec{b}|^2\cos^2\theta.$$

Adding these results gives $$4\,|\vec{a}|^2\,|\vec{b}|^2\bigl(\sin^2\theta + \cos^2\theta\bigr) = 4\,|\vec{a}|^2\,|\vec{b}|^2 = 4 \times 16 \times 9 = 576.$$ Hence, the required value is $$576$$.

We have $$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2$$, $$\vec{a} \cdot \vec{b} = 3$$, and $$|\vec{a} \times \vec{b}|^2 = 75$$.

Expanding the first condition: $$|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2$$. This simplifies to $$2(\vec{a} \cdot \vec{b}) = |\vec{b}|^2$$, so $$|\vec{b}|^2 = 2 \times 3 = 6$$.

Now we use the identity $$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$$. Substituting: $$75 + 9 = |\vec{a}|^2 \times 6$$, so $$|\vec{a}|^2 = \frac{84}{6} = 14$$.

Hence, the correct answer is $$\boxed{14}$$.

We have $$\vec{b} = \hat{i} + \hat{j} + \lambda\hat{k}$$, $$\vec{a} \times \vec{b} = 13\hat{i} - \hat{j} - 4\hat{k}$$, and $$\vec{a} \cdot \vec{b} + 21 = 0$$. Let $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$. Computing $$\vec{a} \times \vec{b}$$ gives $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 1 & 1 & \lambda \end{vmatrix} = (a_2\lambda - a_3)\hat{i} - (a_1\lambda - a_3)\hat{j} + (a_1 - a_2)\hat{k},$$ and equating this with $$13\hat{i} - \hat{j} - 4\hat{k}$$ yields $$a_2\lambda - a_3 = 13 \quad\cdots(1),$$ $$-(a_1\lambda - a_3) = -1 \implies a_3 - a_1\lambda = -1 \quad\cdots(2),$$ $$a_1 - a_2 = -4 \quad\cdots(3).$$

Adding (1) and (2) gives $$(a_2\lambda - a_3)+(a_3 - a_1\lambda)=13+(-1)\implies \lambda(a_2 - a_1)=12.$$ Since (3) gives $$a_2 - a_1=4$$, it follows that $$4\lambda=12$$ and hence $$\lambda=3$$.

The dot product condition $$\vec{a}\cdot\vec{b}=a_1+a_2+3a_3=-21\quad\cdots(4)$$ provides another relation. From (3) we have $$a_1=a_2-4$$, and substituting $$\lambda=3$$ into (1) gives $$3a_2 - a_3=13\implies a_3=3a_2-13$$. Substituting these into (4) yields $$(a_2 - 4)+a_2+3(3a_2 -13)=-21,$$ which simplifies to $$a_2 - 4 + a_2 +9a_2 -39=-21\implies 11a_2 -43=-21\implies 11a_2=22\implies a_2=2.$$ Therefore $$a_1=2-4=-2$$ and $$a_3=3\cdot2-13=-7$$, giving $$\vec{a}=-2\hat{i}+2\hat{j}-7\hat{k}$$ and $$\vec{b}=\hat{i}+\hat{j}+3\hat{k}$$.

It can be verified that $$\vec{a}\times\vec{b}=(2\cdot3-(-7))\hat{i}-((-2)\cdot3-(-7))\hat{j}+((-2)-2)\hat{k}=13\hat{i}-\hat{j}-4\hat{k},$$ and $$\vec{a}\cdot\vec{b}=-2+2-21=-21,$$ so indeed $$\vec{a}\cdot\vec{b}+21=0$$.

Finally, to compute $$(\vec{b}-\vec{a})\cdot(\hat{k}-\hat{j})+(\vec{b}+\vec{a})\cdot(\hat{i}-\hat{k}),$$ note that $$\vec{b}-\vec{a}=(1-(-2))\hat{i}+(1-2)\hat{j}+(3-(-7))\hat{k}=3\hat{i}-\hat{j}+10\hat{k},$$ so $$(\vec{b}-\vec{a})\cdot(\hat{k}-\hat{j})=0+1+10=11.$$ Also, $$\vec{b}+\vec{a}=(1+(-2))\hat{i}+(1+2)\hat{j}+(3+(-7))\hat{k}=-\hat{i}+3\hat{j}-4\hat{k},$$ and $$(\vec{b}+\vec{a})\cdot(\hat{i}-\hat{k})=-1+0+4=3.$$ Therefore the total is $$11+3=14\text{.}$$

The answer is $$\boxed{14}$$.