JEE Three Dimensional Geometry Questions

The line joining $$A(4,7,1)$$ and $$B(3,5,3)$$ will act as the mirror line.
For the image $$Q$$ of $$P(1,0,3)$$ in this line, the line $$AB$$ must be the perpendicular bisector of the segment $$PQ$$. Hence:

1. Find the foot of the perpendicular from $$P$$ to line $$AB$$ (call this foot $$M$$).
2. Use the midpoint condition $$M=\dfrac{P+Q}{2}$$ to get $$Q$$.
3. Add the coordinates of $$Q$$.

Step 1: Equation of line $$AB$$
Direction vector $$\vec d = B-A = (3-4,\,5-7,\,3-1)=(-1,-2,2)$$.

Parametric form: $$\vec r = (4,7,1)+t(-1,-2,2)$$, so any point $$M$$ on the line is $$M(4-t,\;7-2t,\;1+2t)$$.

Step 2: Foot of perpendicular $$M$$ from $$P$$
For $$PM$$ to be perpendicular to $$\vec d$$, we need$$(\vec{M}-\vec{P})\cdot\vec d=0.$$

Compute $$\vec{M}-\vec{P}=(4-t-1,\;7-2t-0,\;1+2t-3)=(3-t,\;7-2t,\;-2+2t).$$

Dot product with $$\vec d=(-1,-2,2):$$
$$(3-t)(-1)+(7-2t)(-2)+(-2+2t)(2)=0.$$

Simplify:
$$-(3-t)-2(7-2t)+2(-2+2t)=0$$
$$(-3+t)+(-14+4t)+(-4+4t)=0$$
$$-21+9t=0 \;\Longrightarrow\; t=\dfrac{21}{9}=\dfrac{7}{3}.$$

Thus $$M\Bigl(4-\dfrac{7}{3},\;7-\dfrac{14}{3},\;1+\dfrac{14}{3}\Bigr)=\Bigl(\dfrac{5}{3},\;\dfrac{7}{3},\;\dfrac{17}{3}\Bigr).$$

Step 3: Coordinates of the image $$Q(\alpha,\beta,\gamma)$$
Since $$M$$ is the midpoint of $$P$$ and $$Q$$,
$$Q = 2M - P.$$

Compute each coordinate:
$$\alpha = 2\!\left(\dfrac{5}{3}\right) - 1 = \dfrac{10}{3}-\dfrac{3}{3}=\dfrac{7}{3},$$ $$\beta = 2\!\left(\dfrac{7}{3}\right) - 0 = \dfrac{14}{3},$$ $$\gamma = 2\!\left(\dfrac{17}{3}\right) - 3 = \dfrac{34}{3}-\dfrac{9}{3}=\dfrac{25}{3}.$$

Step 4: Required sum
$$\alpha+\beta+\gamma = \dfrac{7}{3}+\dfrac{14}{3}+\dfrac{25}{3}=\dfrac{46}{3}.$$

Therefore, $$\alpha + \beta + \gamma = \dfrac{46}{3}$$, which matches Option B.

The given lines can be rewritten in the symmetric-parametric form:
$$L_1 : \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=t \;\;\Longrightarrow\;\; \mathbf{r}_1=(1,2,3)+t\,(2,3,4)$$ $$L_2 : \frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5}=s \;\;\Longrightarrow\;\; \mathbf{r}_2=(\lambda,4,5)+s\,(3,4,5)$$

For two skew lines $$\mathbf{r}=\mathbf{a}_1+t\,\mathbf{b}_1$$ and $$\mathbf{r}=\mathbf{a}_2+s\,\mathbf{b}_2$$, the shortest distance $$D$$ is given by
$$D=\frac{\left|(\mathbf{a}_2-\mathbf{a}_1)\,\cdot\,(\mathbf{b}_1\times\mathbf{b}_2)\right|}{\left|\mathbf{b}_1\times\mathbf{b}_2\right|}$$

Here
$$\mathbf{a}_1=(1,\,2,\,3),\quad \mathbf{b}_1=(2,\,3,\,4)$$ $$\mathbf{a}_2=(\lambda,\,4,\,5),\quad \mathbf{b}_2=(3,\,4,\,5)$$

First find $$\mathbf{b}_1\times\mathbf{b}_2$$:
$$\mathbf{b}_1\times\mathbf{b}_2= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&3&4\\ 3&4&5 \end{vmatrix} =\bigl(3\cdot5-4\cdot4,\;4\cdot3-2\cdot5,\;2\cdot4-3\cdot3\bigr) =(-1,\,2,\,-1)$$ $$\left|\mathbf{b}_1\times\mathbf{b}_2\right|=\sqrt{(-1)^2+2^2+(-1)^2}=\sqrt6$$

Next evaluate $$(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{b}_1\times\mathbf{b}_2)$$:
$$\mathbf{a}_2-\mathbf{a}_1=(\lambda-1,\,4-2,\,5-3)=(\lambda-1,\,2,\,2)$$ $$\bigl(\lambda-1,\,2,\,2\bigr)\cdot(-1,\,2,\,-1)=-(\lambda-1)+4-2=-\lambda+3$$ Hence $$\left|(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{b}_1\times\mathbf{b}_2)\right|=\left|\,\lambda-3\,\right|$$

The shortest distance is given to be $$\dfrac1{\sqrt6}$$, therefore
$$\frac{\left|\lambda-3\right|}{\sqrt6}=\frac1{\sqrt6}\;\;\Longrightarrow\;\;\left|\lambda-3\right|=1$$ Case 1: $$\lambda-3=1\;\Longrightarrow\;\lambda_1=4$$
Case 2: $$\lambda-3=-1\;\Longrightarrow\;\lambda_2=2$$

Thus $$\lambda_1=2,\;\lambda_2=4$$ (order is immaterial).

The circle must pass through the three points
$$A\,(0,0),\quad B\,(\lambda_1,\lambda_2)=(2,4),\quad C\,(\lambda_2,\lambda_1)=(4,2)$$

Compute the side lengths of $$\triangle ABC$$:
$$AB=\sqrt{(2-0)^2+(4-0)^2}=2\sqrt5$$ $$AC=\sqrt{(4-0)^2+(2-0)^2}=2\sqrt5$$ $$BC=\sqrt{(4-2)^2+(2-4)^2}=2\sqrt2$$

Since $$AB=AC$$, the triangle is isosceles. The area $$\Delta$$ can be obtained using the determinant method in 2-D:
$$\Delta=\frac12\left|\,\begin{vmatrix}2&4\\4&2\end{vmatrix}\right| =\frac12\left|\,2\cdot2-4\cdot4\,\right| =\frac12\left|\,4-16\,\right|=6$$

For a triangle with sides $$a,b,c$$ and area $$\Delta$$, the circum-radius $$R$$ is
$$R=\frac{abc}{4\Delta}$$ Taking $$a=BC=2\sqrt2,\; b=AC=2\sqrt5,\; c=AB=2\sqrt5$$:
$$\begin{aligned} R&=\frac{(2\sqrt2)(2\sqrt5)(2\sqrt5)}{4\times6}\\ &=\frac{8\sqrt2\,(5)}{24}\\ &=\frac{40\sqrt2}{24}=\frac{5\sqrt2}{3} \end{aligned}$$

Hence the radius of the required circle is $$\dfrac{5\sqrt2}{3}$$, which corresponds to Option A.

The second line is $$\frac{x}{1} = \frac{y-2}{2} = \frac{z+3}{3}$$, and its direction ratios are (1, 2, 3). A line through (1, 4, 0) with this direction can be written parametrically as $$(1+t, 4+2t, 3t)$$, where $$t$$ is a real parameter.

The first line is given by $$\frac{x-2}{2} = \frac{y-6}{3} = \frac{z-3}{4} = s$$, so any point on it can be expressed as $$(2+2s, 6+3s, 3+4s)$$, where $$s$$ is a real parameter.

$$1+t = 2+2s$$ $$4+2t = 6+3s$$ $$3t = 3+4s$$

From $$1+t = 2+2s$$ we get $$t = 1+2s$$. Substituting into $$4+2t = 6+3s$$ yields $$4+2(1+2s) = 6+3s \Rightarrow 6+4s = 6+3s \Rightarrow s = 0$$, and hence $$t = 1$$. Checking in $$3t = 3+4s$$ gives $$3(1) = 3+4(0) \Rightarrow 3 = 3$$, confirming consistency.

The intersection point is then $$(1+1, 4+2, 3) = (2, 6, 3).$$

The distance from (1, 4, 0) to (2, 6, 3) is $$d = \sqrt{(2-1)^2 + (6-4)^2 + (3-0)^2} = \sqrt{1+4+9} = \sqrt{14}.$$

Given that point $$A(x, y, z)$$ lies in the xy-plane, $$z = 0$$, so $$A = (x, y, 0)$$. It is equidistant from points $$P(0, 3, 2)$$, $$Q(2, 0, 3)$$, and $$R(0, 0, 1)$$.

The distance formula states that the distance between two points $$(x_1, y_1, z_1)$$ and $$(x_2, y_2, z_2)$$ is $$\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}$$.

Setting $$AP = AR$$:

$$\sqrt{(x - 0)^2 + (y - 3)^2 + (0 - 2)^2} = \sqrt{(x - 0)^2 + (y - 0)^2 + (0 - 1)^2}$$

Squaring both sides:

$$x^2 + (y - 3)^2 + 4 = x^2 + y^2 + 1$$

Expanding:

$$x^2 + y^2 - 6y + 9 + 4 = x^2 + y^2 + 1$$

Simplifying:

$$-6y + 13 = 1$$

$$-6y = -12$$

$$y = 2$$

Setting $$AP = AQ$$:

$$\sqrt{(x - 0)^2 + (y - 3)^2 + (0 - 2)^2} = \sqrt{(x - 2)^2 + (y - 0)^2 + (0 - 3)^2}$$

Substituting $$y = 2$$:

$$\sqrt{x^2 + (2 - 3)^2 + 4} = \sqrt{(x - 2)^2 + 2^2 + 9}$$

$$\sqrt{x^2 + 1 + 4} = \sqrt{(x - 2)^2 + 13}$$

$$\sqrt{x^2 + 5} = \sqrt{(x - 2)^2 + 13}$$

Squaring both sides:

$$x^2 + 5 = (x - 2)^2 + 13$$

Expanding:

$$x^2 + 5 = x^2 - 4x + 4 + 13$$

Simplifying:

$$5 = -4x + 17$$

$$4x = 12$$

$$x = 3$$

Thus, $$A = (3, 2, 0)$$.

Given $$B = (1, 4, -1)$$ and $$C = (2, 0, -2)$$, we check the statements.

Statement (S1): $$\triangle ABC$$ is an isosceles right-angled triangle.

Compute the side lengths:

$$AB = \sqrt{(3 - 1)^2 + (2 - 4)^2 + (0 - (-1))^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$

$$AC = \sqrt{(3 - 2)^2 + (2 - 0)^2 + (0 - (-2))^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$

$$BC = \sqrt{(1 - 2)^2 + (4 - 0)^2 + (-1 - (-2))^2} = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$$

Since $$AB = AC = 3$$, $$\triangle ABC$$ is isosceles.

Check Pythagoras theorem:

$$AB^2 + AC^2 = 3^2 + 3^2 = 9 + 9 = 18$$

$$BC^2 = (3\sqrt{2})^2 = 18$$

Since $$AB^2 + AC^2 = BC^2$$, $$\triangle ABC$$ is right-angled at $$A$$. Thus, (S1) is true.

Statement (S2): The area of $$\triangle ABC$$ is $$\frac{9\sqrt{2}}{2}$$.

Since $$\triangle ABC$$ is right-angled at $$A$$, area $$= \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$$.

Alternatively, using vectors:

Vector $$\overrightarrow{AB} = B - A = (1 - 3, 4 - 2, -1 - 0) = (-2, 2, -1)$$

Vector $$\overrightarrow{AC} = C - A = (2 - 3, 0 - 2, -2 - 0) = (-1, -2, -2)$$

Cross product $$\overrightarrow{AB} \times \overrightarrow{AC}$$:

$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 2 & -1 \\ -1 & -2 & -2 \end{vmatrix} = \mathbf{i}(2 \cdot (-2) - (-1) \cdot (-2)) - \mathbf{j}((-2) \cdot (-2) - (-1) \cdot (-1)) + \mathbf{k}((-2) \cdot (-2) - 2 \cdot (-1))$$

$$= \mathbf{i}(-4 - 2) - \mathbf{j}(4 - 1) + \mathbf{k}(4 + 2)$$

$$= -6\mathbf{i} - 3\mathbf{j} + 6\mathbf{k} = (-6, -3, 6)$$

Magnitude $$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-6)^2 + (-3)^2 + 6^2} = \sqrt{36 + 9 + 36} = \sqrt{81} = 9$$

Area $$= \frac{1}{2} \times 9 = \frac{9}{2}$$

But $$\frac{9}{2} \neq \frac{9\sqrt{2}}{2}$$, so (S2) is false.

Therefore, only (S1) is true. The correct option is C.

For the system of equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the system must be consistent.

The coefficient matrix is:

$$A = \begin{bmatrix} \lambda-1 & \lambda-4 & \lambda \\ \lambda & \lambda-1 & \lambda-4 \\ \lambda+1 & \lambda+2 & -(\lambda+2) \end{bmatrix}$$

The determinant of A is computed as:

$$\text{det}(A) = (\lambda-1) \begin{vmatrix} \lambda-1 & \lambda-4 \\ \lambda+2 & -(\lambda+2) \end{vmatrix} - (\lambda-4) \begin{vmatrix} \lambda & \lambda-4 \\ \lambda+1 & -(\lambda+2) \end{vmatrix} + \lambda \begin{vmatrix} \lambda & \lambda-1 \\ \lambda+1 & \lambda+2 \end{vmatrix}$$

Calculate each 2x2 determinant:

First minor: $$\begin{vmatrix} \lambda-1 & \lambda-4 \\ \lambda+2 & -(\lambda+2) \end{vmatrix} = (\lambda-1)(-(\lambda+2)) - (\lambda-4)(\lambda+2) = -(\lambda+2)(2\lambda-5)$$

So, the first term is: $$(\lambda-1) \cdot [-(\lambda+2)(2\lambda-5)] = -(\lambda-1)(\lambda+2)(2\lambda-5)$$

Second minor: $$\begin{vmatrix} \lambda & \lambda-4 \\ \lambda+1 & -(\lambda+2) \end{vmatrix} = \lambda \cdot [-(\lambda+2)] - (\lambda-4)(\lambda+1) = -2\lambda^2 + \lambda + 4$$

So, the second term is: $$- (\lambda-4) \cdot (-2\lambda^2 + \lambda + 4) = -[-2\lambda^3 + 9\lambda^2 - 16] = 2\lambda^3 - 9\lambda^2 + 16$$

Third minor: $$\begin{vmatrix} \lambda & \lambda-1 \\ \lambda+1 & \lambda+2 \end{vmatrix} = \lambda(\lambda+2) - (\lambda-1)(\lambda+1) = 2\lambda + 1$$

So, the third term is: $$\lambda \cdot (2\lambda + 1) = 2\lambda^2 + \lambda$$

Summing all terms: $$\text{det}(A) = -(\lambda-1)(\lambda+2)(2\lambda-5) + 2\lambda^3 - 9\lambda^2 + 16 + 2\lambda^2 + \lambda$$

Expanding $$-(\lambda-1)(\lambda+2)(2\lambda-5)$$: $$(\lambda-1)(\lambda+2) = \lambda^2 + \lambda - 2$$ $$(\lambda^2 + \lambda - 2)(2\lambda - 5) = 2\lambda^3 - 3\lambda^2 - 9\lambda + 10$$ So, $$-(\lambda-1)(\lambda+2)(2\lambda-5) = -2\lambda^3 + 3\lambda^2 + 9\lambda - 10$$

Adding the other terms: $$(-2\lambda^3 + 3\lambda^2 + 9\lambda - 10) + (2\lambda^3 - 9\lambda^2 + 16) + (2\lambda^2 + \lambda) = -4\lambda^2 + 10\lambda + 6$$

Thus, $$\text{det}(A) = -4\lambda^2 + 10\lambda + 6$$.

Set $$\text{det}(A) = 0$$: $$-4\lambda^2 + 10\lambda + 6 = 0$$ Multiply by $$-1$$: $$4\lambda^2 - 10\lambda - 6 = 0$$ Divide by 2: $$2\lambda^2 - 5\lambda - 3 = 0$$

Solve the quadratic equation: $$\lambda = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-3)}}{4} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4}$$ So, $$\lambda = \frac{12}{4} = 3$$ or $$\lambda = \frac{-2}{4} = -\frac{1}{2}$$.

Now, check consistency for each value.

Case $$\lambda = 3$$:

The system becomes: $$2x - y + 3z = 5 \quad \text{(1)}$$ $$3x + 2y - z = 7 \quad \text{(2)}$$ $$4x + 5y - 5z = 9 \quad \text{(3)}$$

Subtract equation (1) from equation (2): $$(3x + 2y - z) - (2x - y + 3z) = 7 - 5 \implies x + 3y - 4z = 2 \quad \text{(4)}$$

Subtract equation (1) from equation (3): $$(4x + 5y - 5z) - (2x - y + 3z) = 9 - 5 \implies 2x + 6y - 8z = 4$$ Divide by 2: $$x + 3y - 4z = 2 \quad \text{(5)}$$

Equations (4) and (5) are identical. The system reduces to: $$2x - y + 3z = 5 \quad \text{and} \quad x + 3y - 4z = 2$$

Solving for $$x$$ and $$y$$ in terms of $$z$$: From equation (5): $$x = 2 - 3y + 4z$$. Substitute into equation (1): $$2(2 - 3y + 4z) - y + 3z = 5 \implies 4 - 6y + 8z - y + 3z = 5 \implies -7y + 11z = 1$$ So, $$y = \frac{11z - 1}{7}$$, and $$x = 2 - 3\left(\frac{11z - 1}{7}\right) + 4z$$. For any $$z$$, there is a solution, so infinitely many solutions.

Case $$\lambda = -\frac{1}{2}$$:

The system becomes: $$-3x - 9y - z = 10 \quad \text{(1)}$$ $$-x - 3y - 9z = 14 \quad \text{(2)}$$ $$x + 3y - 3z = 18 \quad \text{(3)}$$

Add equations (2) and (3): $$(-x - 3y - 9z) + (x + 3y - 3z) = 14 + 18 \implies -12z = 32 \implies z = -\frac{8}{3}$$

Substitute $$z = -\frac{8}{3}$$ into equation (3): $$x + 3y - 3\left(-\frac{8}{3}\right) = 18 \implies x + 3y + 8 = 18 \implies x + 3y = 10$$

Substitute $$z = -\frac{8}{3}$$ into equation (1): $$-3x - 9y - \left(-\frac{8}{3}\right) = 10 \implies -3x - 9y + \frac{8}{3} = 10$$ Multiply by 3: $$-9x - 27y + 8 = 30 \implies -9x - 27y = 22 \quad \text{(4)}$$

From $$x + 3y = 10$$, multiply by $$-9$$: $$-9x - 27y = -90$$. But equation (4) gives $$-9x - 27y = 22$$, contradiction. So inconsistent.

Thus, only $$\lambda = 3$$ gives infinitely many solutions. Now compute $$\lambda^2 + \lambda$$: $$3^2 + 3 = 9 + 3 = 12$$

The answer is 12.

The direction vectors of the given lines are:

For $$L_1$$: $$\vec{d_1} = \langle 1, -1, 2 \rangle$$

For $$L_2$$: $$\vec{d_2} = \langle -1, 2, 1 \rangle$$

The direction vector of $$L_3$$, perpendicular to both, is the cross product:

$$\vec{d_3} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix} = \hat{i}((-1)(1) - (2)(2)) - \hat{j}((1)(1) - (2)(-1)) + \hat{k}((1)(2) - (-1)(-1)) = \hat{i}(-1 - 4) - \hat{j}(1 + 2) + \hat{k}(2 - 1) = \langle -5, -3, 1 \rangle$$

Thus, the direction ratios of $$L_3$$ are $$-5, -3, 1$$.

Parametric equations for $$L_3$$ passing through $$(\alpha, \beta, \gamma)$$ are:

$$x = \alpha - 5t$$
$$y = \beta - 3t$$
$$z = \gamma + t$$

Parametric equations for $$L_1$$ are:

$$x = 1 + s$$
$$y = 2 - s$$
$$z = 1 + 2s$$

Since $$L_3$$ intersects $$L_1$$, set the coordinates equal:

$$1 + s = \alpha - 5t$$   $$-(1)$$
$$2 - s = \beta - 3t$$   $$-(2)$$
$$1 + 2s = \gamma + t$$   $$-(3)$$

Adding equations (1) and (2):

$$(1 + s) + (2 - s) = \alpha - 5t + \beta - 3t$$
$$3 = \alpha + \beta - 8t$$
$$\alpha + \beta - 8t = 3$$   $$-(4)$$

Solving equation (1) for $$s$$:

$$s = \alpha - 5t - 1$$

Substituting into equation (3):

$$1 + 2(\alpha - 5t - 1) = \gamma + t$$
$$1 + 2\alpha - 10t - 2 = \gamma + t$$
$$2\alpha - 10t - 1 = \gamma + t$$
$$2\alpha - \gamma - 11t = 1$$   $$-(5)$$

Eliminating $$t$$ by multiplying equation (4) by 11 and equation (5) by 8:

$$11(\alpha + \beta - 8t) = 11 \times 3$$   ⇒   $$11\alpha + 11\beta - 88t = 33$$   $$-(6)$$
$$8(2\alpha - \gamma - 11t) = 8 \times 1$$   ⇒   $$16\alpha - 8\gamma - 88t = 8$$   $$-(7)$$

Subtracting equation (7) from equation (6):

$$(11\alpha + 11\beta - 88t) - (16\alpha - 8\gamma - 88t) = 33 - 8$$
$$11\alpha + 11\beta - 88t - 16\alpha + 8\gamma + 88t = 25$$
$$-5\alpha + 11\beta + 8\gamma = 25$$

Rearranging:

$$5\alpha - 11\beta - 8\gamma = -25$$

Taking absolute value:

$$|5\alpha - 11\beta - 8\gamma| = |-25| = 25$$

The value is 25.

Let the required line be denoted by $$L$$.
Its symmetric form is given in the question as
$$\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4}\,.$$

Therefore,

 • a point on $$L$$ is $$A(1,\,-4,\,c)$$,
 • its direction vector is $$\vec{n}=(-2,\,d,\,-4)\,.$$

The line $$L$$ is perpendicular to both of the following lines:

$$L_1:\;\vec{r}=\lambda(\hat{i}+a\hat{j}+b\hat{k})\quad\Longrightarrow\quad \text{direction vector }\vec{v_1}=(1,\,a,\,b)$$

$$L_2:\;\vec{r}=(\hat{i}-\hat{j}-6\hat{k})+\mu(-b\hat{i}+a\hat{j}+5\hat{k})\quad\Longrightarrow\quad \text{direction vector }\vec{v_2}=(-b,\,a,\,5)$$

Because $$L$$ is perpendicular to $$L_1$$ and $$L_2$$,

$$\vec{n}\cdot\vec{v_1}=0\quad\text{and}\quad\vec{n}\cdot\vec{v_2}=0\,.$$

Compute each dot product:

1. $$\vec{n}\cdot\vec{v_1}=(-2,\,d,\,-4)\cdot(1,\,a,\,b) =-2+da-4b=0$$ $$\Longrightarrow\;da-4b=2\;-\;(1)$$

2. $$\vec{n}\cdot\vec{v_2}=(-2,\,d,\,-4)\cdot(-b,\,a,\,5) =2b+da-20=0$$ $$\Longrightarrow\;da+2b=20\;-\;(2)$$

The required line $$L$$ must also pass through the point
$$P\Bigl(0,\;-\frac12,\;0\Bigr)$$ given in the statement.

Let the common ratio in the symmetric form be $$t$$. Using point $$A(1,-4,c)$$ and direction $$\vec{n}$$, the parametric equations of $$L$$ are

$$x=1-2t,\qquad y=-4+dt,\qquad z=c-4t\,.$$

Insert point $$P$$:

$$1-2t=0\;\Longrightarrow\;t=\frac12$$

$$-4+dt=-\frac12\;\Longrightarrow\;d\left(\frac12\right)=\frac72 \;\Longrightarrow\;d=7$$

$$c-4t=0\;\Longrightarrow\;c-4\left(\frac12\right)=0 \;\Longrightarrow\;c=2$$

Substitute $$d=7$$ into equations $$(1)$$ and $$(2)$$:

From $$(1):\;7a-4b=2$$

From $$(2):\;7a+2b=20$$

Solve the simultaneous equations:

Add the two equations:
$$\bigl(7a+2b\bigr)+\bigl(7a-4b\bigr)=20+2 \;\Longrightarrow\;14a-2b=22\;-\;(3)$$

Subtract $$(1)$$ from $$(2):\;(7a+2b)-(7a-4b)=20-2 \;\Longrightarrow\;6b=18 \;\Longrightarrow\;b=3$$

Insert $$b=3$$ into $$(1):\;7a-12=2 \;\Longrightarrow\;7a=14 \;\Longrightarrow\;a=2$$

Now we have
$$a=2,\;b=3,\;c=2,\;d=7.$$

Finally,

$$a+b+c+d=2+3+2+7=14.$$

Hence, the value of $$a+b+c+d$$ is $$14$$, which matches Option B.

The perpendicular distance from a point to a line in 3D space is given by the formula:

$$d = \frac{|\overrightarrow{QP} \times \vec{d}|}{|\vec{d}|}$$

where:

• Q is a point on the line,
• $$\vec{d}$$ is the direction vector of the line,
• P is the given point.

First, identify a point on the line. The line is given in symmetric form:

$$\frac{x-1}{2} = \frac{y+2}{-1} = \frac{z+3}{2}$$

Set the parameter $$\lambda = 0$$ to find a point on the line:

When $$\lambda = 0$$:

$$x - 1 = 0 \implies x = 1$$

$$y + 2 = 0 \implies y = -2$$

$$z + 3 = 0 \implies z = -3$$

Thus, point Q is $$(1, -2, -3)$$.

The direction vector $$\vec{d}$$ is given by the denominators: $$(2, -1, 2)$$, so $$\vec{d} = 2\hat{i} - \hat{j} + 2\hat{k}$$.

The given point is P$$(2, -10, 1)$$. The vector $$\overrightarrow{QP}$$ is from Q to P:

$$\overrightarrow{QP} = P - Q = (2 - 1, -10 - (-2), 1 - (-3)) = (1, -8, 4)$$

So, $$\overrightarrow{QP} = \hat{i} - 8\hat{j} + 4\hat{k}$$.

Now, compute the cross product $$\overrightarrow{QP} \times \vec{d}$$:

$$\overrightarrow{QP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \\ \end{vmatrix}$$

Expand the determinant:

$$= \hat{i} \left[ (-8)(2) - (4)(-1) \right] - \hat{j} \left[ (1)(2) - (4)(2) \right] + \hat{k} \left[ (1)(-1) - (-8)(2) \right]$$

Calculate each component:

$$\hat{i}$$ component: $$(-16) - (-4) = -16 + 4 = -12$$

$$\hat{j}$$ component: $$-\left[ 2 - 8 \right] = -[-6] = 6$$ (since the $$\hat{j}$$ term has a negative sign in the expansion)

$$\hat{k}$$ component: $$[-1] - [-16] = -1 + 16 = 15$$

Thus, $$\overrightarrow{QP} \times \vec{d} = -12\hat{i} + 6\hat{j} + 15\hat{k}$$.

Now, find the magnitude:

$$|\overrightarrow{QP} \times \vec{d}| = \sqrt{(-12)^2 + (6)^2 + (15)^2} = \sqrt{144 + 36 + 225} = \sqrt{405}$$

Simplify $$\sqrt{405}$$:

$$\sqrt{405} = \sqrt{81 \times 5} = 9\sqrt{5}$$

Next, find the magnitude of $$\vec{d}$$:

$$|\vec{d}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$

Finally, the perpendicular distance is:

$$d = \frac{|\overrightarrow{QP} \times \vec{d}|}{|\vec{d}|} = \frac{9\sqrt{5}}{3} = 3\sqrt{5}$$

The options are:

A. 6

B. $$5\sqrt{2}$$

C. $$4\sqrt{3}$$

D. $$3\sqrt{5}$$

The distance $$3\sqrt{5}$$ matches option D.

The symmetric form of $$L_1$$ is $$x-1=y-2=z=\lambda$$, so a convenient point on $$L_1$$ is $$A_1(1,2,0)$$ and the direction vector is $$\mathbf{v}=\langle 1,1,1\rangle$$.

For a point $$P(x_0,y_0,z_0)$$, the foot of the perpendicular $$Q$$ on a line through $$A_1$$ with direction $$\mathbf{v}$$ is given by
$$Q=A_1+\dfrac{(P-A_1)\cdot\mathbf{v}}{|\mathbf{v}|^{2}}\;\mathbf{v}$$ $$-(1)$$.
Here $$P(5,1,-3)$$ and $$A_1(1,2,0)$$, so $$P-A_1=\langle 4,-1,-3\rangle$$ and$$(P-A_1)\cdot\mathbf{v}=4+(-1)+(-3)=0.$$
Using $$(1)$$ with numerator $$0$$ gives $$Q=A_1=(1,2,0).$$ Thus $$Q(1,2,0).$$

The symmetric form of $$L_2$$ is $$x-2=y=z-1=\mu$$, so a point on $$L_2$$ is $$A_2(2,0,1)$$ with the same direction $$\mathbf{v}=\langle 1,1,1\rangle$$.

Applying $$-(1)$$ to $$L_2$$: $$P-A_2=\langle 3,1,-4\rangle$$ and$$(P-A_2)\cdot\mathbf{v}=3+1+(-4)=0.$$
Hence $$R=A_2=(2,0,1).$$ Thus $$R(2,0,1).$$

Compute the vectors of the sides of $$\triangle PQR$$:
$$\overrightarrow{PQ}=Q-P=\langle -4,1,3\rangle,$$
$$\overrightarrow{PR}=R-P=\langle -3,-1,4\rangle.$$

The area of a triangle whose sides are $$\mathbf{a}$$ and $$\mathbf{b}$$ is $$\dfrac{1}{2}\,|\mathbf{a}\times\mathbf{b}|$$. First find the cross product:
$$\overrightarrow{PQ}\times\overrightarrow{PR}=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ -4&\;1&\;3\\ -3&-1&\;4\end{vmatrix}$$
$$=\mathbf{i}(1\cdot4-3\cdot(-1))-\mathbf{j}((-4)\cdot4-3\cdot(-3))+\mathbf{k}((-4)\cdot(-1)-1\cdot(-3))$$
$$=\mathbf{i}(4+3)-\mathbf{j}(-16+9)+\mathbf{k}(4+3)$$
$$=\langle 7,7,7\rangle.$$

Magnitude of the cross product:$$|\langle 7,7,7\rangle|=\sqrt{7^{2}+7^{2}+7^{2}}=\sqrt{147}=7\sqrt{3}.$$

Therefore, the area of $$\triangle PQR$$ is
$$A=\dfrac{1}{2}\,(7\sqrt{3})=\dfrac{7}{2}\sqrt{3}.$$

Finally, $$4A^{2}=4\left(\dfrac{7}{2}\sqrt{3}\right)^{2}=4\left(\dfrac{49\cdot3}{4}\right)=147.$$

Hence $$4A^{2}=147$$, which corresponds to Option B.

Line L passes through $$P(2,-1,3)$$ and is perpendicular to:

$$L_1$$: direction $$(2,1,-2)$$

$$L_2$$: direction $$(1,3,4)$$

Direction of L = $$(2,1,-2) \times (1,3,4)$$:

$$ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix} = \hat{i}(4+6) - \hat{j}(8+2) + \hat{k}(6-1) = (10, -10, 5) $$

Simplifying: direction $$(2, -2, 1)$$.

Line L: $$(x,y,z) = (2,-1,3) + t(2,-2,1)$$

To find intersection with yz-plane ($$x = 0$$):

$$2 + 2t = 0 \Rightarrow t = -1$$

$$Q = (0, -1+2, 3-1) = (0, 1, 2)$$

Distance PQ:

$$PQ = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2} = \sqrt{4+4+1} = \sqrt{9} = 3$$

The correct answer is Option 4: 3.

The line having both Q and R can be written in parametric form as
$$(x,y,z)=\left(5t-3,\;2t+1,\;3t-4\right),\quad t\in\mathbb{R}\; -(1)$$

Let $$t=a$$ give point $$Q$$ and $$t=b$$ give point $$R$$:
$$Q\;(5a-3,\;2a+1,\;3a-4),\qquad R\;(5b-3,\;2b+1,\;3b-4)$$

Set $$k=b-a$$. Then the vector $$\overrightarrow{QR}$$ is
$$\overrightarrow{QR}=(5k,\;2k,\;3k)$$

Its length is given to be $$5$$:
$$|\overrightarrow{QR}|=\lvert k\rvert\sqrt{5^{2}+2^{2}+3^{2}} =\lvert k\rvert\sqrt{38}=5$$
$$\Rightarrow\;\lvert k\rvert=\dfrac{5}{\sqrt{38}}\; -(2)$$

Point $$P$$ is fixed at $$(0,2,3)$$. Write the vectors from $$P$$ to $$Q$$ and $$R$$:
$$\overrightarrow{PQ}=Q-P=(5a-3,\;2a-1,\;3a-7)$$
$$\overrightarrow{PR}=R-P=\overrightarrow{PQ}+\overrightarrow{QR}$$

The area of $$\triangle PQR$$ is
$$\text{Area}=\dfrac12\, \lvert\overrightarrow{PQ}\times\overrightarrow{PR}\rvert$$

Because $$\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}$$, and $$\overrightarrow{PQ}\times\overrightarrow{PQ}=0$$, we have
$$\overrightarrow{PQ}\times\overrightarrow{PR} =\overrightarrow{PQ}\times\overrightarrow{QR}$$

Compute the cross-product. Put
$$\overrightarrow{PQ}=(5a-3,\;2a-1,\;3a-7),\; \overrightarrow{QR}=(5k,\;2k,\;3k)$$

$$$ \begin{aligned} \overrightarrow{PQ}\times\overrightarrow{QR} &=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 5a-3 & 2a-1 & 3a-7\\ 5k & 2k & 3k \end{vmatrix}\\[4pt] &=k\bigl(11,\,-26,\,-1\bigr) \end{aligned} $$$

$$\lvert\overrightarrow{PQ}\times\overrightarrow{QR}\rvert =|k|\sqrt{11^{2}+(-26)^{2}+(-1)^{2}} =|k|\sqrt{121+676+1}=|k|\sqrt{798}$$

Using $$-(2)$$, $$|k|=\dfrac{5}{\sqrt{38}}$$, hence
$$\text{Area}= \dfrac12\cdot\dfrac{5}{\sqrt{38}}\cdot\sqrt{798} =\dfrac{5}{2}\sqrt{\dfrac{798}{38}} =\dfrac{5}{2}\sqrt{21}\; -(3)$$

Thus $$\text{Area}=\dfrac{5\sqrt{21}}{2}$$. If the area is written as $$\dfrac{m}{n}$$, comparison with $$-(3)$$ gives
$$\dfrac{m}{n}=\dfrac{5\sqrt{21}}{2} \;\Longrightarrow\;2m-5\sqrt{21}\,n=0$$

Therefore the relation between $$m$$ and $$n$$ is
Option B: $$2m-5\sqrt{21}n=0$$

We start by parametrizing the ray from $$P$$ in the given direction. A point on the ray is: $$ Q = P + t\vec{d} = \left(\frac{15}{7}+t,\; \frac{32}{7}+4t,\; 7+7t\right) $$.

Next, the line can be parametrized as $$(x,y,z) = (-1+3s,\; -3+5s,\; -5+7s)$$. Setting $$Q$$ equal to a point on this line gives

$$ \frac{15}{7} + t = -1 + 3s \quad \cdots(1) $$

$$ \frac{32}{7} + 4t = -3 + 5s \quad \cdots(2) $$

$$ 7 + 7t = -5 + 7s \quad \cdots(3) $$

From (3) we obtain $$7s - 7t = 12 \implies s - t = \frac{12}{7} \implies s = t + \frac{12}{7} \quad \cdots(4)$$. Substituting this into (1) yields

$$\frac{15}{7} + t = -1 + 3\left(t + \frac{12}{7}\right) = -1 + 3t + \frac{36}{7}$$

$$\frac{15}{7} + t = \frac{-7+36}{7} + 3t = \frac{29}{7} + 3t$$

$$\frac{15}{7} - \frac{29}{7} = 2t$$

$$-\frac{14}{7} = 2t \implies t = -1$$

It follows from (4) that $$s = -1 + \frac{12}{7} = \frac{5}{7}$$.

As a check, substituting into (2) gives on the left $$\frac{32}{7} + 4(-1) = \frac{32}{7} - 4 = \frac{4}{7}$$ and on the right $$-3 + 5 \cdot \frac{5}{7} = -3 + \frac{25}{7} = \frac{4}{7}$$, confirming consistency.

The distance from $$P$$ to the intersection point $$Q$$ equals $$|t| \cdot |\vec{d}|$$, where $$|t| = 1$$ and $$|\vec{d}| = \sqrt{1+16+49} = \sqrt{66}$$. Therefore, the square of the distance is

$$ |t|^2 \cdot |\vec{d}|^2 = 1 \times 66 = 66 $$.

The correct answer is Option 4: 66.

Let the lengths of the mutually perpendicular edges $$AB$$, $$AC$$, and $$AD$$ be $$x$$, $$y$$, and $$z$$ respectively.

Step 1: Express the given face areas in terms of edge lengths

The areas of the right-angled triangles are given by:

$$\text{Area}(\triangle ABC) = \frac{1}{2}xy = 5 \implies xy = 10$$

$$\text{Area}(\triangle ACD) = \frac{1}{2}yz = 6 \implies yz = 12$$

$$\text{Area}(\triangle ADB) = \frac{1}{2}zx = 7 \implies zx = 14$$

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Step 2: Relate the tetrahedron along the 3D plane

Let vertex $$A$$ be the origin $$(0,0,0)$$. Then the coordinates of the vertices are:

$$A(0, 0, 0), \quad B(x, 0, 0), \quad C(0, y, 0), \quad D(0, 0, z)$$

We form two side vectors for triangle $$BCD$$ from vertex $$B$$:

$$\vec{BC} = -x\hat{i} + y\hat{j}$$

$$\vec{BD} = -x\hat{i} + z\hat{k}$$

The cross product $$\vec{BC} \times \vec{BD}$$ is evaluated using a determinant:

$$\vec{BC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -x & y & 0 \\ -x & 0 & z \end{vmatrix} = (yz)\hat{i} + (zx)\hat{j} + (xy)\hat{k}$$

The area of triangle $$BCD$$ is half the magnitude of this cross product vector:

$$\text{Area}(\triangle BCD) = \frac{1}{2}\sqrt{(yz)^2 + (zx)^2 + (xy)^2}$$

$$\text{Area}(\triangle BCD) = \sqrt{\left(\frac{1}{2}yz\right)^2 + \left(\frac{1}{2}zx\right)^2 + \left(\frac{1}{2}xy\right)^2}$$

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Step 3: Substitute the known area values

Substituting the given face areas into the derived relation:

$$\text{Area}(\triangle BCD) = \sqrt{6^2 + 7^2 + 5^2}$$

$$\text{Area}(\triangle BCD) = \sqrt{36 + 49 + 25} = \sqrt{110}$$

Write the two lines in parametric form.

For $$L_1$$, put parameter $$t$$:
$$\frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}\; \Longrightarrow \; x=7+t,\; y=5,\; z=3-t$$

For $$L_2$$, put parameter $$s$$:
$$\frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}\; \Longrightarrow \; x=1+3s,\; y=-3+4s,\; z=-7+5s$$

To find the point of intersection A, equate the coordinates:

$$7+t = 1+3s,\quad 5 = -3+4s,\quad 3-t = -7+5s$$

From $$5 = -3+4s$$ we get $$4s=8$$, so $$s=2$$.

Substituting $$s=2$$ in the $$x$$-equation gives $$7+t = 1+6$$, hence $$t=0$$.
The $$z$$-equation is also satisfied. Therefore

$$A(7,\,5,\,3).$$

Next locate point $$B$$ on $$L_1$$. Let its parameter be $$t=k$$:

$$B(7+k,\,5,\,3-k).$$

Distance formula gives
$$AB^2 = (k)^2 + 0^2 + (-k)^2 = 2k^2.$$ Given $$AB = \sqrt{15}$$, so

$$2k^2 = 15 \;\Longrightarrow\; k^2 = \frac{15}{2}. \;-(1)$$

Similarly, point $$C$$ lies on $$L_2$$. Write $$s = 2 + \ell$$:

$$C(7+3\ell,\,5+4\ell,\,3+5\ell).$$

Distance AC:
$$AC^2 = (3\ell)^2 + (4\ell)^2 + (5\ell)^2 = 9\ell^2 +16\ell^2 +25\ell^2 = 50\ell^2.$$ Given $$AC = \sqrt{15}$$, so

$$50\ell^2 = 15 \;\Longrightarrow\; \ell^2 = \frac{3}{10}. \;-(2)$$

Take any sign choices for $$k$$ and $$\ell$$; the area depends on $$k^2$$ and $$\ell^2$$ only.

Find vectors $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$:

$$\overrightarrow{AB} = (k,\,0,\,-k),\quad \overrightarrow{AC} = (3\ell,\,4\ell,\,5\ell).$$

The cross product is

$$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\[2pt] k&0&-k\\[2pt] 3\ell&4\ell&5\ell \end{vmatrix} = (4k\ell)\,\mathbf{i} \;-\; (8k\ell)\,\mathbf{j} \;+\; (4k\ell)\,\mathbf{k}.$$ Thus
$$\lvert\overrightarrow{AB}\times\overrightarrow{AC}\rvert^2 = (4k\ell)^2(1^2+(-2)^2+1^2) = 16k^2\ell^2 \times 6 = 96k^2\ell^2.$$

Area of triangle ABC is
$$\frac{1}{2}\lvert\overrightarrow{AB}\times\overrightarrow{AC}\rvert,$$ so the square of the area equals

$$\left(\frac{1}{2}\right)^2 \times 96k^2\ell^2 = 24k^2\ell^2.$$ Insert $$k^2$$ from $$(1)$$ and $$\ell^2$$ from $$(2)$$:

$$24 \times \frac{15}{2} \times \frac{3}{10} = 24 \times \frac{45}{20} = 24 \times \frac{9}{4} = 6 \times 9 = 54.$$

Therefore, the square of the area of $$\triangle ABC$$ is $$54$$.
Hence Option A is correct.

We need to find the distance of the point $$P(7, 10, 11)$$ from line $$L_1: \dfrac{x-4}{1} = \dfrac{y-4}{0} = \dfrac{z-2}{3}$$ along line $$L_2: \dfrac{x-9}{2} = \dfrac{y-13}{3} = \dfrac{z-17}{6}$$.

The line through $$P(7, 10, 11)$$ parallel to $$L_2$$ (direction $$(2, 3, 6)$$) is: $$\dfrac{x-7}{2} = \dfrac{y-10}{3} = \dfrac{z-11}{6} = t$$

A general point on this line is $$(7 + 2t, 10 + 3t, 11 + 6t)$$.

This point must lie on $$L_1$$. From $$L_1$$: $$y = 4$$ (since the direction ratio for $$y$$ is 0).

So $$10 + 3t = 4$$, giving $$t = -2$$.

The point of intersection is $$(7 - 4, 4, 11 - 12) = (3, 4, -1)$$.

Let us verify this lies on $$L_1$$: $$\dfrac{3-4}{1} = -1$$ and $$\dfrac{-1-2}{3} = -1$$. Yes, it does.

The distance from $$P(7, 10, 11)$$ to $$(3, 4, -1)$$ is:

$$d = \sqrt{(7-3)^2 + (10-4)^2 + (11-(-1))^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$$

Hence, the correct answer is Option B.

Here is the mathematically formatted version of your solution using standard LaTeX:

Given that the vector $\mathbf{a}$ can be resolved into a component along vector $$\mathbf{b}$$ and a component perpendicular to vector $$\mathbf{b}$$:

$$\mathbf{a} = \mathbf{a}_{\parallel} + \mathbf{a}_{\perp}$$

$$\mathbf{a}$$ = $$\frac{16}{11}(3\hat{i} + \hat{j} - \hat{k})$$ + $$\frac{1}{11}(-4\hat{i} - 5\hat{j} - 17\hat{k})$$

Taking $$\frac{1}{11}$$ as a common factor:

$$\mathbf{a} = \frac{1}{11} \left[ (48\hat{i} + 16\hat{j} - 16\hat{k}) + (-4\hat{i} - 5\hat{j} - 17\hat{k}) \right]$$

$$\mathbf{a} = \frac{1}{11} (44\hat{i} + 11\hat{j} - 33\hat{k})$$ $$\mathbf{a} = 4\hat{i} + \hat{j} - 3\hat{k}$$

Let $$\mathbf{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$$. By comparing coefficients, we get:

• $$\alpha = 4$$
• $$\beta = 1$$
• $$\gamma = -3$$

Now, calculate $$\alpha^2 + \beta^2 + \gamma^2$$:

$$\alpha^2 + \beta^2 + \gamma^2 = (4)^2 + (1)^2 + (-3)^2$$ $$\alpha^2 + \beta^2 + \gamma^2 = 16 + 1 + 9 = 26$$

The given lines are
  $$L_1:\;\dfrac{x-1}{2}= \dfrac{y-2}{3}= \dfrac{z-3}{4}$$
  $$L_2:\;x-6 = y = -z+4$$

Write them in parametric form.

For $$L_1$$ let the parameter be $$t$$:
  $$x = 1+2t,\;y = 2+3t,\;z = 3+4t\;.\tag{-1}$$

Hence the intersection point with $$L_1$$ is
  $$A(\alpha,\beta,\gamma)=\bigl(1+2t,\;2+3t,\;3+4t\bigr).$$

For $$L_2$$ write $$y=\lambda$$. From $$x-6=y$$ we get $$x = 6+\lambda$$, and from $$y = -z+4$$ we get $$z = 4-\lambda$$. So

$$B(a,b,c)=\bigl(6+\lambda,\;\lambda,\;4-\lambda\bigr).\tag{-2}$$

The required line passes through the given point $$P(4,1,0)$$ and the two points $$A$$ and $$B$$, so the three points are collinear. Therefore the vectors $$\overrightarrow{PA}$$ and $$\overrightarrow{PB}$$ must be parallel.

Compute the two vectors:

$$\overrightarrow{PA} = (\,(1+2t)-4,\;(2+3t)-1,\;(3+4t)-0) = (-3+2t,\;1+3t,\;3+4t).$$

$$\overrightarrow{PB} = (\,(6+\lambda)-4,\;\lambda-1,\;(4-\lambda)-0) = (2+\lambda,\;\lambda-1,\;4-\lambda).$$

Parallelism means the three component ratios are equal to a common constant $$k$$:

$$\dfrac{-3+2t}{2+\lambda}= \dfrac{1+3t}{\,\lambda-1\,}= \dfrac{3+4t}{\,4-\lambda\,}=k.\tag{-3}$$

First equate the ratios formed by the first two components:

$$( -3+2t)(\lambda-1) = (1+3t)(2+\lambda).\tag{-4}$$

Next equate the ratios formed by the first and the third components:

$$( -3+2t)(4-\lambda) = (3+4t)(2+\lambda).\tag{-5}$$

Expand $$( -4)$$:

$$( -3+2t)\lambda -(-3+2t) = (1+3t)2 + (1+3t)\lambda$$
$$(-3+2t)\lambda +3 -2t = 2 + 6t + (1+3t)\lambda.$$

Collecting terms gives $$(-4-t)\lambda +1 -8t = 0.$$ Hence

$$\lambda(-4-t)=8t-1.\tag{-6}$$

Expand $$( -5)$$:

$$( -3+2t)(4-\lambda) = (3+4t)(2+\lambda)$$
$$-12 + 8t +3\lambda -2t\lambda = 6 + 8t + 3\lambda + 4t\lambda.$$

Simplifying yields $$-18 - 6t\lambda = 0 \quad\Longrightarrow\quad t\lambda = -3.\tag{-7}$$

Use $$( -7)$$ in $$( -6)$$. From $$( -7)$$, $$\lambda= -\dfrac{3}{t} \;(t\neq 0).$$ Substitute in $$( -6)$$:

$$\left(-\dfrac{3}{t}\right)(-4-t) = 8t-1$$
$$\dfrac{12}{t} + 3 = 8t - 1.$$

Multiply by $$t$$:

$$12 + 3t = 8t^{2} - t.$$

Re-arrange:

$$8t^{2} -4t -12 = 0 \quad\Longrightarrow\quad 2t^{2}-t-3=0.$$

Quadratic formula gives $$t = \dfrac{1\pm 5}{4}\;\Longrightarrow\; t=\dfrac{3}{2}\;\text{or}\;t=-1.$$

Check each value in $$( -3)$$ to ensure a single proportionality constant $$k$$ exists.

$$A(-1,-1,-1),\qquad B(9,3,1).$$

Vectors $$\overrightarrow{PA}=(-5,\; -2,\; -1),\quad \overrightarrow{PB}=(5,\; 2,\; 1)$$ are negatives of each other, so the points $$P,A,B$$ are indeed collinear. Thus $$t=-1,\;\lambda=3$$ is the only acceptable solution.

Fill the matrix with the obtained coordinates:

$$\begin{vmatrix} 1 & 0 & 1\\ \alpha & \beta & \gamma\\ a & b & c \end{vmatrix} = \begin{vmatrix} 1 & 0 & 1\\ -1 & -1 & -1\\ 9 & 3 & 1 \end{vmatrix}.$$

Evaluate the determinant:

$$\begin{aligned} \Delta &= 1\bigl((-1)(1) - (-1)(3)\bigr) - 0(\cdots) + 1\bigl((-1)(3) - (-1)(9)\bigr)\\[4pt] &= 1(-1 + 3) + 1(-3 + 9)\\ &= 1(2) + 1(6)\\ &= 8. \end{aligned}$$

Hence $$\begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix}=8.$$

The correct choice is Option A (8).

 $$P=(6,7,7)$$, on the line direction vector is $$\vec{d} = (3,2,-2)$$. The vector from $$P$$ to $$B$$ is $$\vec{PB} = (1-6, 2-7, 3-7) = (-5,-5,-4)\,.$$

The area of the parallelogram determined by $$\vec{PB}$$ and $$\vec{d}$$ is the magnitude of their cross product:

$$\vec{PB} \times \vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-5&-5&-4\\3&2&-2\end{vmatrix} = \hat{i}(10+8) - \hat{j}(10+12) + \hat{k}(-10+15) = (18, -22, 5)\,,$$

so $$|\vec{PB} \times \vec{d}| = \sqrt{324+484+25} = \sqrt{833}$$. Since $$|\vec{d}| = \sqrt{9+4+4} = \sqrt{17}\,,$$ the distance from B to the line is

$$\frac{\sqrt{833}}{\sqrt{17}} = \sqrt{\frac{833}{17}} = \sqrt{49} = 7\,,$$

which is the length of the altitude.

Finally, the area of triangle ABC is half the product of its base AC and its height:

$$\text{Area} = \frac{1}{2} \times AC \times h = \frac{1}{2} \times 6 \times 7 = 21\,.$$

Line $$L_1$$ passes through $$P_1(1,2,3)$$ and is parallel to the $$z$$-axis.
Hence a direction vector of $$L_1$$ is $$\mathbf{b}=(0,0,1)$$ and its equation is $$x=1,\;y=2,\;z=t$$.

Line $$L_2$$ passes through $$P_2(\lambda,5,6)$$ and is parallel to the $$y$$-axis.
Its direction vector is $$\mathbf{d}=(0,1,0)$$ and its equation is $$x=\lambda,\;y=5+s,\;z=6$$.

The shortest (perpendicular) distance $$D$$ between two skew lines $$\mathbf{r}=\mathbf{a}+t\mathbf{b}$$ and $$\mathbf{r}=\mathbf{c}+s\mathbf{d}$$ is given by
$$D=\frac{\lvert(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}\times\mathbf{d})\rvert}{\lvert\mathbf{b}\times\mathbf{d}\rvert}$$.

First find $$\mathbf{b}\times\mathbf{d}$$:
$$\mathbf{b}\times\mathbf{d}=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 0&0&1\\ 0&1&0 \end{vmatrix}=(-1,0,0).$$
Its magnitude is $$\lvert\mathbf{b}\times\mathbf{d}\rvert=1$$.

Next, $$\mathbf{a}-\mathbf{c}=P_1P_2=(1-\lambda,\,2-5,\;3-6)=(1-\lambda,-3,-3)$$.

Dot product:
$$(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}\times\mathbf{d})=(1-\lambda,-3,-3)\cdot(-1,0,0)=\lambda-1.$$

Therefore the shortest distance is
$$D=\lvert\lambda-1\rvert.$$

The question states this distance equals $$3$$, so
$$\lvert\lambda-1\rvert=3 \;\Longrightarrow\; \lambda=4 \text{ or } \lambda=-2.$$
Thus $$\lambda_1=4,\;\lambda_2=-2 \;(\lambda_2\lt\lambda_1).$$

The required point is $$(\lambda_1,\lambda_2,7)=(4,-2,7).$$ Let this point be $$Q(4,-2,7).$$

Distance of a point $$Q$$ from a line through $$P_1$$ with direction $$\mathbf{b}$$ is
$$\text{dist}=\frac{\lvert\overrightarrow{P_1Q}\times\mathbf{b}\rvert}{\lvert\mathbf{b}\rvert}.$$

Compute $$\overrightarrow{P_1Q}=(4-1,\,-2-2,\,7-3)=(3,-4,4).$$

Cross product:
$$\overrightarrow{P_1Q}\times\mathbf{b}=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 3&-4&4\\ 0&0&1 \end{vmatrix}=(-4,-3,0).$$

Magnitude:
$$\lvert(-4,-3,0)\rvert=\sqrt{(-4)^2+(-3)^2}= \sqrt{16+9}=5.$$

Because $$\lvert\mathbf{b}\rvert=1$$, the perpendicular distance is $$5$$.

Hence the square of this distance is $$5^2=25$$.

Therefore the required value is $$25$$, i.e. Option C.

Parametrizing $$L$$ as $$(1+2t, -1+3t, 4t)$$ and substituting $$1 + 2t = 5$$ gives $$t = 2$$, which yields $$p = -1 + 6 = 5$$ and $$q = 8$$, so that $$R = (5,5,8)$$.

To find the image $$P$$ of $$Q(7,-2,5)$$ in the line, let the foot of the perpendicular from $$Q$$ onto $$L$$ be $$M = (1+2t, -1+3t, 4t)$$ so that $$\vec{QM} = (2t-6, 3t+1, 4t-5)$$ is perpendicular to the direction vector $$(2,3,4)$$ of $$L$$. Imposing $$2(2t-6) + 3(3t+1) + 4(4t-5) = 0$$ gives $$4t-12 + 9t+3 + 16t-20 = 0$$, hence $$29t - 29 = 0 \Rightarrow t = 1$$, and thus $$M = (3,2,4)$$.

Since $$P$$ is the reflection of $$Q$$ across the line, we have $$M = \frac{P+Q}{2}$$ and therefore $$P = 2M - Q = (6-7,4-(-2),8-5) = (-1,6,3)$$.

Next, we compute the area of triangle $$PQR$$ by forming the vectors $$\vec{QP} = P - Q = (-1-7,6-(-2),3-5) = (-8,8,-2)$$ and $$\vec{QR} = R - Q = (5-7,5-(-2),8-5) = (-2,7,3)$$.

Their cross product is given by $$\vec{QP} \times \vec{QR} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -8 & 8 & -2 \\ -2 & 7 & 3 \end{vmatrix}$$ which equals $$= \vec{i}(24-(-14)) - \vec{j}(-24-4) + \vec{k}(-56-(-16))$$, simplifies to $$= \vec{i}(24+14) - \vec{j}(-28) + \vec{k}(-56+16)$$, and hence $$= 38\vec{i} + 28\vec{j} - 40\vec{k}$$. This gives $$|\vec{QP} \times \vec{QR}|^2 = 38^2 + 28^2 + 40^2 = 1444 + 784 + 1600 = 3828$$. 

Since the area is $$\frac{1}{2}|\vec{QP} \times \vec{QR}|$$, its square is $$(\text{Area})^2 = \frac{1}{4} \times 3828 = 957$$.

To find the area $$A$$ of the triangle, we first need to determine its vertices by finding the points of intersection between the three given lines.

Given three lines are:

Line 1 ($$L_1$$): $$x + 2 = y - 1 = z$$

Line 2 ($$L_2$$): $$\frac{x-3}{5} = \frac{y}{-1} = \frac{z-1}{1}$$

Line 3 ($$L_3$$): $$\frac{x}{-3} = \frac{y-3}{3} = \frac{z-2}{1}$$

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Step 1: Finding Intersection of $$L_1$$ and $$L_2$$. Let it be A 

From $$L_1$$, we can express $$x$$ and $$y$$ in terms of $$z$$:

$$x = z - 2$$

$$y = z + 1$$

Substituting these into the equation of $$L_2$$:

$$\frac{(z - 2) - 3}{5} = \frac{z + 1}{-1} = \frac{z - 1}{1}$$

Comparing the last two parts:

$$\frac{z + 1}{-1} = z - 1 \implies -z - 1 = z - 1 \implies 2z = 0 \implies z = 0$$

Substituting $$z = 0$$ back gives:

$$x = 0 - 2 = -2$$

$$y = 0 + 1 = 1$$

Thus, the first vertex is $$A(-2, 1, 0)$$.

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Step 2: Finding Intersection of $$L_1$$ and $$L_3$$. Let it be B

Using the same expressions from $$L_1$$ ($$x = z - 2$$ and $$y = z + 1$$), substitute them into the equation of $$L_3$$:

$$\frac{z - 2}{-3} = \frac{(z + 1) - 3}{3} = \frac{z - 2}{1}$$

Comparing the first and third parts:

$$\frac{z - 2}{-3} = \frac{z - 2}{1} \implies z - 2 = -3(z - 2) \implies 4(z - 2) = 0 \implies z = 2$$

Substituting $$z = 2$$ back gives:

$$x = 2 - 2 = 0$$

$$y = 2 + 1 = 3$$

Thus, the second vertex is $$B(0, 3, 2)$$.

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Step 3: Finding Intersection of $$L_2$$ and $$L_3$$. Let it be C

Let a general point on $$L_2$$ be parameterized by $$t$$:

$$x = 5t + 3, \quad y = -t, \quad z = t + 1$$

Substitute these coordinates into the equation of $$L_3$$:

$$\frac{5t + 3}{-3} = \frac{-t - 3}{3} = \frac{(t + 1) - 2}{1}$$

$$\frac{5t + 3}{-3} = \frac{-t - 3}{3} = t - 1$$

Comparing the second and third parts:

$$\frac{-t - 3}{3} = t - 1 \implies -t - 3 = 3t - 3 \implies 4t = 0 \implies t = 0$$

Substituting $$t = 0$$ into the parametric coordinates gives:

$$x = 3, \quad y = 0, \quad z = 1$$

Thus, the third vertex is $$C(3, 0, 1)$$.

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Step 4: Calculate the Area $$A$$ using Vectors

Now form two adjacent side vectors from vertex $$A$$:

$$\vec{AB} = (0 - (-2))\hat{i} + (3 - 1)\hat{j} + (2 - 0)\hat{k} = 2\hat{i} + 2\hat{j} + 2\hat{k}$$

$$\vec{AC} = (3 - (-2))\hat{i} + (0 - 1)\hat{j} + (1 - 0)\hat{k} = 5\hat{i} - \hat{j} + \hat{k}$$

Find the cross product $$\vec{AB} \times \vec{AC}$$:

$$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 2 \\ 5 & -1 & 1 \end{vmatrix}$$

$$= \hat{i}(2(1) - 2(-1)) - \hat{j}(2(1) - 2(5)) + \hat{k}(2(-1) - 2(5))$$

$$= 4\hat{i} + 8\hat{j} - 12\hat{k}$$

The area $$A$$ of the triangle is half the magnitude of this cross product vector:

$$A = \frac{1}{2} \sqrt{4^2 + 8^2 + (-12)^2}$$

$$A = \frac{1}{2} \sqrt{16 + 64 + 144} = \frac{1}{2} \sqrt{224}$$

Squaring the area gives:

$$A^2 = \left(\frac{1}{2} \sqrt{224}\right)^2 = \frac{1}{4} \times 224 = 56$$

We have $$L_1: \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$$ and $$L_2: \frac{x-2}{2} = \frac{y}{0} = \frac{z+4}{\alpha}$$.

Find the intersection point $$B$$.

Parametric forms: $$L_1: (1+3t,\; 1-t,\; -1)$$ and $$L_2: (2+2s,\; 0,\; -4+\alpha s)$$.

Setting coordinates equal:

From $$y$$: $$1 - t = 0 \implies t = 1$$

From $$x$$: $$1 + 3(1) = 2 + 2s \implies s = 1$$

From $$z$$: $$-1 = -4 + \alpha(1) \implies \alpha = 3$$

So $$B = (4, 0, -1)$$ and $$\alpha = 3$$.

Find foot of perpendicular $$P$$ from $$A(1, 1, -1)$$ on $$L_2$$.

A general point on $$L_2$$: $$(2 + 2\delta,\; 0,\; -4 + 3\delta)$$.

$$\vec{AP} = (1 + 2\delta,\; -1,\; -3 + 3\delta)$$

Direction of $$L_2$$: $$(2, 0, 3)$$.

For perpendicularity: $$\vec{AP} \cdot (2, 0, 3) = 0$$

$$2(1 + 2\delta) + 0 + 3(-3 + 3\delta) = 0$$

$$2 + 4\delta - 9 + 9\delta = 0 \implies 13\delta = 7 \implies \delta = \frac{7}{13}$$

$$P = \left(\frac{40}{13},\; 0,\; -\frac{31}{13}\right)$$

Compute $$(PB)^2$$.

$$\vec{PB} = \left(4 - \frac{40}{13},\; 0,\; -1 + \frac{31}{13}\right) = \left(\frac{12}{13},\; 0,\; \frac{18}{13}\right)$$

$$(PB)^2 = \frac{144}{169} + \frac{324}{169} = \frac{468}{169}$$

Calculate $$26\alpha(PB)^2$$.

$$26 \times 3 \times \frac{468}{169} = \frac{36504}{169} = 216$$

The answer is $$\boxed{216}$$.

The line passes through point P(-2, -1, 3) and is parallel to the vector $$3\widehat{i} + 2\widehat{j} + 2\widehat{k}$$. Therefore, the parametric equations for any point on the line are:

$$x = -2 + 3t$$
$$y = -1 + 2t$$
$$z = 3 + 2t$$

where $$t$$ is a scalar parameter.

Let Q be a point on this line, so Q has coordinates $$(-2 + 3t, -1 + 2t, 3 + 2t)$$. The distance from Q to R(1, 3, 3) is given as 5. Using the distance formula:

$$\sqrt{ [(-2 + 3t) - 1]^2 + [(-1 + 2t) - 3]^2 + [(3 + 2t) - 3]^2 } = 5$$

Simplify the expressions inside the square root:

$$(-2 + 3t) - 1 = 3t - 3$$
$$(-1 + 2t) - 3 = 2t - 4$$
$$(3 + 2t) - 3 = 2t$$

So the equation becomes:

$$\sqrt{ (3t - 3)^2 + (2t - 4)^2 + (2t)^2 } = 5$$

Square both sides to eliminate the square root:

$$(3t - 3)^2 + (2t - 4)^2 + (2t)^2 = 25$$

Expand each term:

$$(3t - 3)^2 = 9t^2 - 18t + 9$$
$$(2t - 4)^2 = 4t^2 - 16t + 16$$
$$(2t)^2 = 4t^2$$

Sum them up:

$$9t^2 - 18t + 9 + 4t^2 - 16t + 16 + 4t^2 = 25$$
$$(9t^2 + 4t^2 + 4t^2) + (-18t - 16t) + (9 + 16) = 25$$
$$17t^2 - 34t + 25 = 25$$

Subtract 25 from both sides:

$$17t^2 - 34t = 0$$

Factor out 17t:

$$17t(t - 2) = 0$$

So, $$t = 0$$ or $$t = 2$$.

When $$t = 0$$, Q is $$(-2 + 3(0), -1 + 2(0), 3 + 2(0)) = (-2, -1, 3)$$, which is point P. However, if Q coincides with P, the triangle PQR degenerates to a line, so we discard this solution.

When $$t = 2$$, Q is $$(-2 + 3(2), -1 + 2(2), 3 + 2(2)) = (4, 3, 7)$$.

Thus, the points are P(-2, -1, 3), Q(4, 3, 7), and R(1, 3, 3). To find the area of triangle PQR, use the formula for the area of a triangle given three points in space. The area is half the magnitude of the cross product of vectors $$\overrightarrow{PQ}$$ and $$\overrightarrow{PR}$$.

First, find the vectors:

$$\overrightarrow{PQ} = Q - P = (4 - (-2), (3 - (-1)), (7 - 3)) = (6, 4, 4)$$
$$\overrightarrow{PR} = R - P = (1 - (-2)), (3 - (-1)), (3 - 3)) = (3, 4, 0)$$

Now, compute the cross product $$\overrightarrow{PQ} \times \overrightarrow{PR}$$:

$$\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0 \\ \end{vmatrix} = \widehat{i}(4 \cdot 0 - 4 \cdot 4) - \widehat{j}(6 \cdot 0 - 4 \cdot 3) + \widehat{k}(6 \cdot 4 - 4 \cdot 3)$$

Calculate each component:

$$\widehat{i}(0 - 16) = -16\widehat{i}$$
$$-\widehat{j}(0 - 12) = -\widehat{j}(-12) = 12\widehat{j}$$
$$\widehat{k}(24 - 12) = 12\widehat{k}$$

So, the cross product vector is $$(-16, 12, 12)$$.

The magnitude of this vector is:

$$\sqrt{(-16)^2 + 12^2 + 12^2} = \sqrt{256 + 144 + 144} = \sqrt{544}$$

Simplify $$\sqrt{544}$$:

$$544 = 16 \times 34$$
$$\sqrt{544} = \sqrt{16 \times 34} = 4\sqrt{34}$$

Thus, the area of triangle PQR is:

$$\frac{1}{2} \times 4\sqrt{34} = 2\sqrt{34}$$

The square of the area is:

$$(2\sqrt{34})^2 = 4 \times 34 = 136$$

Therefore, the square of the area of $$\triangle PQR$$ is 136.

The equations of the two given planes are
  $$\pi_1: \; 2x + 3y + z = 4,$$
  $$\pi_2: \; x + 2y + z = 5.$$

1. Direction vector of their line of intersection $$L_1$$
For two intersecting planes the direction vector is the cross-product of their normals.
  Normal of $$\pi_1 : \; \mathbf{n}_1 = (2,3,1)$$
  Normal of $$\pi_2 : \; \mathbf{n}_2 = (1,2,1)$$
$$\mathbf{d}= \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 3 & 1\\ 1 & 2 & 1 \end{vmatrix} = (1,\,-1,\,1).$$

2. One point on $$L_1$$
Put $$z=0$$ in both plane equations:
$$2x + 3y = 4, \qquad x + 2y = 5.$$ Solving, $$y=6,\; x=-7.$$
So $$A(-7,\,6,\,0)$$ lies on $$L_1.$$ Hence $$L_1:\; (x,y,z)=(-7,6,0)+t(1,-1,1).$$

3. Equation of $$L_2$$
$$L_2$$ passes through $$P(2,-1,3)$$ and is parallel to $$L_1,$$ so it has the same direction vector $$\mathbf{d}=(1,-1,1).$$
$$L_2:\; (x,y,z)=(2,-1,3)+t(1,-1,1).$$

4. Point $$Q$$ where $$L_2$$ meets plane $$M: 2x + y - 2z = 6$$
Substitute $$x=2+t,\; y=-1-t,\; z=3+t$$ into the plane:
$$2(2+t)+(-1-t)-2(3+t)=6$$ $$4+2t-1-t-6-2t=6$$ $$-3 - t = 6 \;\Longrightarrow\; t=-9.$$ Therefore $$Q:(x,y,z)=\bigl(2-9,\,-1+9,\,3-9\bigr)=(-7,\,8,\,-6).$$

5. Length of segment $$PQ$$
$$\overrightarrow{PQ}=Q-P=(-9,\,9,\,-9).$$
$$|PQ|=\sqrt{(-9)^2+9^2+(-9)^2}=9\sqrt{3}.$$ Option A is TRUE.

6. Foot $$R$$ of the perpendicular from $$P$$ to plane $$M$$
Plane $$M: 2x + y - 2z - 6 = 0$$ has normal $$\mathbf{n}=(2,1,-2).$$
For a point $$P(x_0,y_0,z_0)$$, the foot is $$R = P - \frac{ax_0 + by_0 + cz_0 + d}{a^2+b^2+c^2}\,(a,b,c).$$ Here $$ax_0+by_0+cz_0+d = 2(2)+1(-1)+(-2)(3)-6=-9,$$ $$a^2+b^2+c^2=4+1+4=9,$$ so the factor is $$\dfrac{-9}{9}=-1.$$ Hence $$R = (2,-1,3) - (-1)(2,1,-2) = (2+2,\,-1+1,\,3+(-2)) = (4,\,0,\,1).$$

7. Length of segment $$QR$$
$$\overrightarrow{QR}=R-Q=(11,\,-8,\,7).$$
$$|QR|=\sqrt{11^2+(-8)^2+7^2}=\sqrt{234}\;(\approx 15.329).$$ Option B states the length is 15, so it is FALSE.

8. Area of $$\triangle PQR$$
Using vectors $$\overrightarrow{PQ}=(-9,9,-9)$$ and $$\overrightarrow{PR}=R-P=(2,1,-2),$$
$$\overrightarrow{PQ}\times\overrightarrow{PR} =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ -9&9&-9\\ 2&1&-2 \end{vmatrix} =(-9,\,-36,\,-27).$$ Magnitude $$= \sqrt{(-9)^2+(-36)^2+(-27)^2} = \sqrt{2106}=3\sqrt{234}.$$ Area $$\triangle PQR=\frac12\bigl| \overrightarrow{PQ}\times\overrightarrow{PR}\bigr| =\frac12\,(3\sqrt{234})=\frac{3}{2}\sqrt{234}.$$ Option C is TRUE.

9. Acute angle between $$PQ$$ and $$PR$$
$$\overrightarrow{PQ}\cdot\overrightarrow{PR}=(-9)(2)+9(1)+(-9)(-2)=9.$$ $$|PQ|=9\sqrt3,\quad |PR|=\sqrt{2^2+1^2+(-2)^2}=3.$$ $$\cos\theta = \frac{9}{9\sqrt3\cdot3}= \frac{1}{3\sqrt3}.$$ Option D gives $$\cos^{-1}\!\left(\dfrac{1}{2\sqrt3}\right),$$ hence it is FALSE.

Conclusion
Correct statements: Option A and Option C.

The first line is given by $$\frac{x-2}{1} = \frac{y-1}{2} = \frac{z+3}{-3}$$, with point $$(2, 1, -3)$$ and direction vector $$\vec{d_1}=(1,2,-3)$$

The second line is given by $$\frac{x+1}{2} = \frac{y+3}{4} = \frac{z+5}{-5}$$, with point $$(-1, -3, -5)$$ and direction vector $$\vec{d_2} = (2, 4, -5)$$.

Since $$(1,2,-3)$$ is not a scalar multiple of $$(2,4,-5)$$ , the lines are skew.

 $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}  = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = 2\hat{i} - \hat{j} + 0\hat{k} = (2, -1, 0)$$.

The vector joining the given points on the two lines is $$\vec{P_1P_2} = (-1-2, -3-1, -5+3) = (-3, -4, -2)$$.

The shortest distance between skew lines is given by $$d = \frac{|\vec{P_1P_2} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}\,. $$ Here $$\vec{P_1P_2} \cdot (2, -1, 0) = (-3)(2) + (-4)(-1) + (-2)(0) = -6 + 4 + 0 = -2$$ and $$|\vec{d_1} \times \vec{d_2}| = \sqrt{4+1+0} = \sqrt{5}\,, $$ so $$d = \frac{|-2|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$$.

Squaring this distance yields $$d^2 = \frac{4}{5}\,, $$ so $$m = 4$$, $$n = 5$$ are coprime and $$m + n = 4 + 5 = 9\$$

The first line can be written in vector form as
$$\mathbf{r}= \mathbf{a}_1 + t\,\mathbf{b}_1,$$
where a point on the line is $$\mathbf{a}_1 = (3,\; \alpha,\; 3)$$ and the direction vector is $$\mathbf{b}_1 = \langle 3,\; -1,\; 1\rangle.$$

The second line is
$$\mathbf{r}= \mathbf{a}_2 + s\,\mathbf{b}_2,$$
with $$\mathbf{a}_2 = (-3,\; -7,\; \beta)$$ and $$\mathbf{b}_2 = \langle -3,\; 2,\; 4\rangle.$$

For two skew lines the shortest distance is given by
$$D = \frac{\left| (\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{b}_1 \times \mathbf{b}_2) \right|}{\left| \mathbf{b}_1 \times \mathbf{b}_2 \right|}. \qquad -(1)$$

Compute the cross product $$\mathbf{b}_1 \times \mathbf{b}_2$$:

$$ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 3 & -1 & 1\\ -3 & 2 & 4 \end{vmatrix} = \langle (-1)(4)-1(2),\; -(3\cdot 4-1\cdot(-3)),\; 3\cdot 2-(-1)(-3) \rangle = \langle -6,\; -15,\; 3\rangle. $$

Thus $$\mathbf{b}_1 \times \mathbf{b}_2 = \langle -6,\; -15,\; 3\rangle = -3\langle 2,\; 5,\; -1\rangle.$$

Magnitude of this vector:
$$\left| \mathbf{b}_1 \times \mathbf{b}_2 \right| = 3\sqrt{2^{2}+5^{2}+(-1)^{2}} = 3\sqrt{30}. \qquad -(2)$$

Vector joining the two given points:
$$(\mathbf{a}_2-\mathbf{a}_1) = \langle -3-3,\; -7-\alpha,\; \beta-3\rangle = \langle -6,\; -7-\alpha,\; \beta-3\rangle.$$

Now evaluate the scalar triple product in the numerator of (1):

$$ (\mathbf{a}_2-\mathbf{a}_1)$$\cdot$$(\mathbf{b}_1 $$\times$$ \mathbf{b}_2) = \langle -6,\; -7-$$\alpha$$,\; $$\beta$$-3\rangle $$\cdot$$ \langle -6,\; -15,\; 3\rangle $$ $$ = (-6)(-6) + (-7-$$\alpha$$)(-15) + ($$\beta$$-3)(3) $$ $$ = 36 + 15(7+$$\alpha$$) + 3($$\beta$$-3) $$ $$ = 36 + 105 + 15$$\alpha$$ + 3$$\beta$$ - 9 $$ $$ = 132 + 15$$\alpha$$ + 3$$\beta$$. \qquad -(3) $$

Given shortest distance $$D = 3$$\sqrt{30}$$,$$ substitute (2) and (3) into (1):

$$ $$\frac{\left| 132 + 15\alpha + 3\beta \right|}{3\sqrt{30}$$} = 3$$\sqrt{30}$$. $$

Multiply both sides by $$3$$\sqrt{30}$$:$$
$$  $$\left$$| 132 + 15$$\alpha$$ + 3$$\beta$$ $$\right$$| = 9 $$\times$$ 30 = 270. \qquad -(4)$$

Equation (4) gives two possibilities:
$$132 + 15$$\alpha$$ + 3$$\beta$$ = 270 \quad$$\text{or}$$\quad 132 + 15$$\alpha$$ + 3$$\beta$$ = -270.$$ Divide each by $$3$$:

Case 1: $$5$$\alpha + \beta$$ = 46.$$
Case 2: $$5$$\alpha + \beta$$ = -134.$$

The problem asks for the positive value of $$5$$\alpha + \beta$$$$, hence we choose
Case 1: $$5$$\alpha + \beta$$ = 46.$$

Therefore, the required positive value is $$\mathbf{46}$$, which corresponds to Option B.

For the line $$\dfrac{x+1}{3}=\dfrac{y}{4}=\dfrac{z}{5}$$ put the common parameter $$t$$:
  $$x=-1+3t,\;y=0+4t,\;z=0+5t$$
Hence a point on the line is $$A(-1,\,0,\,0)$$ and its direction vector is $$\vec{d_1}= \langle 3,\,4,\,5\rangle$$.

The other line is given in vector form
  $$\vec{r}= (p\hat i+2\hat j+\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$$
so a point on it is $$B(p,\,2,\,1)$$ and its direction vector is $$\vec{d_2}= \langle 2,\,3,\,4\rangle$$.

Formula for shortest distance between two skew lines:
  $$D=\frac{\left|(\overrightarrow{AB}\cdot(\vec{d_1}\times\vec{d_2}))\right|}{\left|\vec{d_1}\times\vec{d_2}\right|}$$
where $$\overrightarrow{AB}=B-A=\langle p+1,\,2,\,1\rangle$$.

First evaluate the cross-product:
  $$\vec{d_1}\times\vec{d_2}= \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] 3 & 4 & 5\\ 2 & 3 & 4 \end{vmatrix} =\langle 1,\,-2,\,1\rangle$$
  $$|\vec{d_1}\times\vec{d_2}|=\sqrt{1^{2}+(-2)^{2}+1^{2}}=\sqrt6$$.

Triple scalar product:
  $$\overrightarrow{AB}\cdot(\vec{d_1}\times\vec{d_2}) =\langle p+1,\,2,\,1\rangle\cdot\langle 1,\,-2,\,1\rangle =(p+1)(1)+2(-2)+1(1)=p-2$$.

Given shortest distance $$D=\dfrac{1}{\sqrt6}$$, therefore
  $$\frac{|p-2|}{\sqrt6}=\frac{1}{\sqrt6}\;\;\Longrightarrow\;\;|p-2|=1$$
  $$\Rightarrow\;p=1\;\text{or}\;p=3$$.

Arranging $$a

The required ellipse is
  $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \;\Longrightarrow\; \frac{x^{2}}{1^{2}}+\frac{y^{2}}{3^{2}}=1$$
so the semi-major axis is $$b_{\max}=3$$ and the semi-minor axis is $$a_{\min}=1$$.

For an ellipse, length of the latus rectum $$L$$ is
  $$L=\frac{2\,(\,\text{(semi-minor)}\,)^{2}}{\text{(semi-major)}}$$
Hence
  $$L=\frac{2\,(1)^{2}}{3}=\frac{2}{3}$$.

Therefore the length of the latus rectum is $$\dfrac{2}{3}$$, which is Option C.

Let $$L_1$$ pass through $$A(7,\,6,\,2)$$ with direction vector $$\vec{a}= -3\hat{i}+2\hat{j}+4\hat{k}$$ and $$L_2$$ pass through $$B(5,\,3,\,4)$$ with direction vector $$\vec{b}= 2\hat{i}+\hat{j}+3\hat{k}$$.

The shortest distance $$d$$ between two skew lines is given by
$$d=\frac{\bigl|(\vec{AB}\,.\,(\vec{a}\times\vec{b}))\bigr|}{\lVert\vec{a}\times\vec{b}\rVert}$$
where $$\vec{AB}=\overrightarrow{BA}$$ or $$\overrightarrow{AB}$$ is the vector joining any point on one line to any point on the other.

Step 1 : Vector joining the two reference points
$$\vec{AB}= \bigl(5-7\bigr)\hat{i}+\bigl(3-6\bigr)\hat{j}+\bigl(4-2\bigr)\hat{k}= -2\hat{i}-3\hat{j}+2\hat{k}$$ $$-(1)$$

Step 2 : Cross product of the direction vectors
$$\vec{a}\times\vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ -3 & 2 & 4\\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(2\cdot3-4\cdot1)\;-\;\hat{j}((-3)\cdot3-4\cdot2)\;+\;\hat{k}((-3)\cdot1-2\cdot2)$$ $$= 2\hat{i}+17\hat{j}-7\hat{k}$$ $$-(2)$$

Step 3 : Magnitude of the cross product
$$\lVert\vec{a}\times\vec{b}\rVert= \sqrt{2^{2}+17^{2}+(-7)^{2}} =\sqrt{4+289+49} =\sqrt{342} =3\sqrt{38}$$ $$-(3)$$

Step 4 : Scalar triple product
$$(\vec{AB}\,.\,(\vec{a}\times\vec{b}))= (-2)(2)+(-3)(17)+(2)(-7) =-4-51-14 =-69$$
Hence $$\bigl|(\vec{AB}\,.\,(\vec{a}\times\vec{b}))\bigr|=69$$ $$-(4)$$

Step 5 : Shortest distance
Substituting $$(3)$$ and $$(4)$$ into the formula, we get
$$d=\frac{69}{3\sqrt{38}} =\frac{23}{\sqrt{38}}$$ $$-(5)$$

Thus the shortest distance between $$L_1$$ and $$L_2$$ is $$\dfrac{23}{\sqrt{38}}$$.

Option A is correct.

A line through (−1,2,1) parallel to the direction (2,3,4) intersects the line $$\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$$ at P. To find P, we parametrize the first line as $$(-1+2s,\,2+3s,\,1+4s)$$ and the second line as $$(-2+3t,\,3+2t,\,4+t)\,$$.

Equating the coordinates yields the system:
$$-1+2s=-2+3t\quad\Rightarrow\quad2s-3t=-1$$
$$2+3s=3+2t\quad\Rightarrow\quad3s-2t=1$$
$$1+4s=4+t\quad\Rightarrow\quad4s-t=3$$.

From the second equation, $$t=\frac{3s-1}{2}\,. $$ Substituting into the first gives
$$2s-3\cdot\frac{3s-1}{2}=-1\quad\Rightarrow\quad4s-9s+3=-2\quad\Rightarrow\quad-5s=-5\quad\Rightarrow\quad s=1,$$
so $$t=\frac{3(1)-1}{2}=1\,. $$ Checking the third equation confirms $$4(1)-1=3\,. $$

Hence $$P=(-1+2\,,\,2+3\,,\,1+4)=(1,5,5)\,. $$ The distance from P to $$Q(4,-5,1)$$ is $$\sqrt{(4-1)^2+(-5-5)^2+(1-5)^2}=\sqrt{9+100+16}=\sqrt{125}=5\sqrt{5}\,. $$

The correct answer is Option 2: $$5\sqrt{5}$$.

Given lines $$L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ and $$L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$.

Point on $$L_1$$: $$A = (1, 2, 3)$$, direction $$\vec{d_1} = (2, 3, 4)$$.

Point on $$L_2$$: $$B = (2, 4, 5)$$, direction $$\vec{d_2} = (3, 4, 5)$$.

The shortest distance line is perpendicular to both $$L_1$$ and $$L_2$$, so its direction is $$\vec{d_1} \times \vec{d_2}$$:

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = (15-16)\hat{i} - (10-12)\hat{j} + (8-9)\hat{k} = (-1, 2, -1)$$

Let the point on $$L_1$$ be $$P = (1+2t, 2+3t, 3+4t)$$ and on $$L_2$$ be $$Q = (2+3s, 4+4s, 5+5s)$$.

$$\vec{PQ}$$ must be parallel to $$(-1, 2, -1)$$:

$$\vec{PQ} = (1+3s-2t, 2+4s-3t, 2+5s-4t)$$

Setting $$\vec{PQ} \cdot \vec{d_1} = 0$$:

$$2(1+3s-2t) + 3(2+4s-3t) + 4(2+5s-4t) = 0$$

$$2+6s-4t+6+12s-9t+8+20s-16t = 0$$

$$16 + 38s - 29t = 0 \quad \cdots (1)$$

Setting $$\vec{PQ} \cdot \vec{d_2} = 0$$:

$$3(1+3s-2t) + 4(2+4s-3t) + 5(2+5s-4t) = 0$$

$$3+9s-6t+8+16s-12t+10+25s-20t = 0$$

$$21 + 50s - 38t = 0 \quad \cdots (2)$$

From (1): $$29t = 16 + 38s \implies t = \frac{16+38s}{29}$$

Substituting into (2): $$21 + 50s - 38 \cdot \frac{16+38s}{29} = 0$$

$$29(21+50s) - 38(16+38s) = 0$$

$$609 + 1450s - 608 - 1444s = 0$$

$$1 + 6s = 0 \implies s = -\frac{1}{6}$$

$$t = \frac{16 + 38(-1/6)}{29} = \frac{16 - 19/3}{29} = \frac{29/3}{29} = \frac{1}{3}$$

$$P = \left(1+\frac{2}{3},\; 2+1,\; 3+\frac{4}{3}\right) = \left(\frac{5}{3}, 3, \frac{13}{3}\right)$$

$$Q = \left(2-\frac{1}{2},\; 4-\frac{2}{3},\; 5-\frac{5}{6}\right) = \left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)$$

The line of shortest distance passes through $$P = \left(\frac{5}{3}, 3, \frac{13}{3}\right)$$ with direction $$(-1, 2, -1)$$:

$$\text{Parametric form: } \left(\frac{5}{3} - \lambda,\; 3 + 2\lambda,\; \frac{13}{3} - \lambda\right)$$

Checking Option (1): $$\left(\frac{14}{3}, -3, \frac{22}{3}\right)$$:

$$\frac{5}{3} - \lambda = \frac{14}{3} \implies \lambda = -3$$

$$3 + 2(-3) = 3 - 6 = -3 \quad \checkmark$$

$$\frac{13}{3} - (-3) = \frac{13}{3} + 3 = \frac{22}{3} \quad \checkmark$$

The correct answer is Option (1): $$\boxed{\left(\frac{14}{3}, -3, \frac{22}{3}\right)}$$.

We are given $$P(-1, -1, 2)$$, $$Q(5, 5, 10)$$, and $$O$$ is the origin.

Point $$A$$ divides $$PQ$$ internally in the ratio $$r : 1$$. By the section formula:

$$\overrightarrow{OA} = \frac{r \cdot \overrightarrow{OQ} + 1 \cdot \overrightarrow{OP}}{r + 1} = \frac{r(5, 5, 10) + (-1, -1, 2)}{r + 1} = \left(\frac{5r - 1}{r + 1}, \frac{5r - 1}{r + 1}, \frac{10r + 2}{r + 1}\right)$$

We have $$\overrightarrow{OQ} = (5, 5, 10)$$ and $$\overrightarrow{OP} = (-1, -1, 2)$$.

Computing $$\overrightarrow{OQ} \cdot \overrightarrow{OA}$$:

$$\overrightarrow{OQ} \cdot \overrightarrow{OA} = \frac{5(5r-1) + 5(5r-1) + 10(10r+2)}{r+1} = \frac{25r - 5 + 25r - 5 + 100r + 20}{r+1} = \frac{150r + 10}{r+1}$$

The problem involves $$|\overrightarrow{OP} \times \overrightarrow{OA}|^2$$ (cross product). Computing $$\overrightarrow{OP} \times \overrightarrow{OA}$$:

$$\overrightarrow{OP} \times \overrightarrow{OA} = \frac{1}{r+1} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 2 \\ 5r-1 & 5r-1 & 10r+2 \end{vmatrix}$$

$$\hat{i}$$-component: $$(-1)(10r+2) - 2(5r-1) = -10r - 2 - 10r + 2 = -20r$$

$$\hat{j}$$-component: $$-((-1)(10r+2) - 2(5r-1)) = -(-10r - 2 - 10r + 2) = 20r$$

$$\hat{k}$$-component: $$(-1)(5r-1) - (-1)(5r-1) = 0$$

$$\overrightarrow{OP} \times \overrightarrow{OA} = \frac{1}{r+1}(-20r, 20r, 0) = \frac{20r}{r+1}(-1, 1, 0)$$

$$|\overrightarrow{OP} \times \overrightarrow{OA}|^2 = \frac{400r^2}{(r+1)^2} \times (1 + 1 + 0) = \frac{800r^2}{(r+1)^2}$$

Substituting into the given equation $$(\overrightarrow{OQ} \cdot \overrightarrow{OA}) - \frac{1}{5}|\overrightarrow{OP} \times \overrightarrow{OA}|^2 = 10$$:

$$\frac{150r + 10}{r+1} - \frac{1}{5} \cdot \frac{800r^2}{(r+1)^2} = 10$$

$$\frac{150r + 10}{r+1} - \frac{160r^2}{(r+1)^2} = 10$$

Multiplying through by $$(r+1)^2$$:

$$(150r + 10)(r+1) - 160r^2 = 10(r+1)^2$$

$$150r^2 + 150r + 10r + 10 - 160r^2 = 10r^2 + 20r + 10$$

$$-10r^2 + 160r + 10 = 10r^2 + 20r + 10$$

$$0 = 20r^2 - 140r$$

$$0 = 20r(r - 7)$$

Since $$r > 0$$, we get $$r = 7$$.

The answer is Option D: $$7$$.

 First, we observe that the line passes through $$A(1, 2, 1)$$ with direction vector $$\vec{d} = (2, 1, 3)$$, so a general point on the line can be written as $$Q = (1 + 2t,\; 2 + t,\; 1 + 3t)$$.

Next, we form the vector from this point to $$P$$, namely $$\overrightarrow{QP} = P - Q = (4 - 1 - 2t,\; 4 - 2 - t,\; 3 - 1 - 3t) = (3 - 2t,\; 2 - t,\; 2 - 3t)$$. 

Since the foot of the perpendicular from $$P$$ to the line occurs when this vector is orthogonal to the direction vector $$\vec{d}$$, we impose $$\overrightarrow{QP}\cdot \vec{d} = 0$$, which gives $$2(3 - 2t) + 1(2 - t) + 3(2 - 3t) = 0,$$ leading to $$6 - 4t + 2 - t + 6 - 9t = 14 - 14t = 0$$ and hence $$t = 1$$.

Substituting $$t = 1$$ back into the parametric form yields the foot of the perpendicular $$F = (1 + 2(1),\; 2 + 1,\; 1 + 3(1)) = (3, 3, 4).$$

Finally, the reflection of $$P$$ across this foot is calculated by $$(\alpha, \beta, \gamma) = 2F - P = 2(3, 3, 4) - (4, 4, 3) = (6 - 4,\; 6 - 4,\; 8 - 3) = (2, 2, 5),$$ and thus $$\alpha + \beta + \gamma = 2 + 2 + 5 = 9.$$

The given line $$L$$ can be written in parametric form as
$$x = 6 + 3t,\; y = 7 + 2t,\; z = 7 - 2t$$
so its direction vector is $$\mathbf{d} = \langle 3,\,2,\,-2\rangle$$.

Let the point $$P(1,\,2,\,3)$$ drop a perpendicular to $$L$$ at the foot $$H(6+3t,\,7+2t,\,7-2t)$$.
For the foot of the perpendicular, $$\overrightarrow{PH} \cdot \mathbf{d}=0$$.

Compute $$\overrightarrow{PH}$$:
$$\overrightarrow{PH}= \langle 6+3t-1,\; 7+2t-2,\; 7-2t-3\rangle =\langle 5+3t,\; 5+2t,\; 4-2t\rangle.$$

Form the dot product and set it to zero:
$$(5+3t)\cdot 3 + (5+2t)\cdot 2 + (4-2t)\cdot (-2) = 0.$$

Simplify:
$$3(5+3t) + 2(5+2t) -2(4-2t)=0$$
$$15+9t + 10+4t -8+4t = 0$$
$$17 + 17t = 0$$
$$\Rightarrow \; t = -1.$$

Thus
$$H = \bigl(6+3(-1),\,7+2(-1),\,7-2(-1)\bigr) = (3,\,5,\,9).$$

The magnitude of the direction vector is
$$|\mathbf{d}|=\sqrt{3^{2}+2^{2}+(-2)^{2}}=\sqrt{9+4+4}=\sqrt{17}.$$

Two points $$A$$ and $$B$$ on $$L$$ are each at distance $$2\sqrt{17}$$ from $$H$$.
If their parameters are $$t_1$$ and $$t_2$$ then
$$|t_1+1|\cdot|\mathbf{d}| = 2\sqrt{17},\qquad |t_2+1|\cdot|\mathbf{d}| = 2\sqrt{17}.$$

Since $$|\mathbf{d}|=\sqrt{17}$$, this gives
$$|t_1+1| = 2,\quad |t_2+1| = 2 \; \Longrightarrow\; t_1 = -3,\; t_2 = 1.$$

Coordinates of the two points:

$$t=-3 \Longrightarrow A(-3,\,1,\,13).$$

$$t=1 \Longrightarrow B(9,\,9,\,5).$$

Vectors from the origin are therefore
$$\vec{OA}=\langle -3,\,1,\,13\rangle,\qquad \vec{OB}=\langle 9,\,9,\,5\rangle.$$

The required dot product is
$$\vec{OA}\cdot\vec{OB} = (-3)(9) + (1)(9) + (13)(5) = -27 + 9 + 65 = 47.$$

Hence $$\vec{OA}\cdot\vec{OB}=47$$.
Option B is correct.

Direction ratios of $$L_1$$ are $$\langle 1,2,3\rangle$$ and those of $$L_2$$ are $$\langle 3,2,\gamma\rangle$$.
Write the two lines in the symmetric-to-parametric form.

For $$L_1$$ let the parameter be $$\lambda$$:
$$x=-11+\lambda,\;y=-21+2\lambda,\;z=-29+3\lambda$$

For $$L_2$$ let the parameter be $$\mu$$:
$$x=-16+3\mu,\;y=-11+2\mu,\;z=-4+\gamma\mu$$

Because the lines intersect, there exist $$\lambda,\mu$$ for which the coordinates coincide:

$$-11+\lambda=-16+3\mu\;-(1)$$
$$-21+2\lambda=-11+2\mu\;-(2)$$
$$-29+3\lambda=-4+\gamma\mu\;-(3)$$

Solve $$(1)$$ and $$(2)$$ first.
From $$(2):\;2\lambda-2\mu=10\Rightarrow\lambda-\mu=5\;-(4)$$
From $$(1):\;\lambda-3\mu=-5\;-(5)$$
Substitute $$\lambda=\mu+5$$ from $$(4)$$ into $$(5)$$:
$$\mu+5-3\mu=-5\;\Longrightarrow\;-2\mu=-10\;\Longrightarrow\;\mu=5$$
Hence $$\lambda=\mu+5=10$$.

Coordinates of the point of intersection $$R_1$$ (use $$L_1$$):
$$x=-11+10=-1,\;y=-21+20=-1,\;z=-29+30=1$$
Therefore $$\overrightarrow{OR_1}=\langle-1,-1,1\rangle$$.

Determine $$\gamma$$ using $$(3)$$:
Left side: $$-29+3\lambda=-29+30=1$$
Right side: $$-4+\gamma\mu=-4+5\gamma$$
Equating: $$1=-4+5\gamma\;\Longrightarrow\;5\gamma=5\;\Longrightarrow\;\gamma=1$$.

Now find a normal to the plane containing the two lines.
Two non-parallel direction vectors are
$$\mathbf{v}_1=\langle1,2,3\rangle,\qquad\mathbf{v}_2=\langle3,2,1\rangle$$ (because $$\gamma=1$$).
Take the cross product:

$$\mathbf{n}_0=\mathbf{v}_1\times\mathbf{v}_2 =\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 1&2&3\\ 3&2&1 \end{vmatrix} =\langle-4,\,8,\,-4\rangle=-4\langle1,-2,1\rangle$$

An associated unit normal vector is therefore
$$\hat{n}=\frac{1}{\sqrt6}\langle1,-2,1\rangle =\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}$$.

Finally, compute the scalar product $$\overrightarrow{OR_1}\cdot\hat{n}$$:

$$\langle-1,-1,1\rangle\cdot\frac{1}{\sqrt6}\langle1,-2,1\rangle =\frac{1}{\sqrt6}\,[-1(1)+(-1)(-2)+1(1)] =\frac{1}{\sqrt6}\,( -1+2+1)=\frac{2}{\sqrt6} =\sqrt{\frac{2}{3}}$$

Summarising our results:
$$\gamma=1,\quad\overrightarrow{OR_1}=\,-\hat{i}-\hat{j}+\hat{k},\quad \hat{n}=\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k},\quad \overrightarrow{OR_1}\cdot\hat{n}=\sqrt{\frac{2}{3}}$$

Matching with List-II:
(P) $$\gamma$$ → (3) $$1$$
(Q) $$\hat{n}$$ → (4) $$\frac{1}{\sqrt{6}}\hat{i}-\frac{2}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$$
(R) $$\overrightarrow{OR_1}$$ → (1) $$-\hat{i}-\hat{j}+\hat{k}$$
(S) $$\overrightarrow{OR_1}\cdot\hat{n}$$ → (5) $$\sqrt{\frac{2}{3}}$$

Thus the correct option is
Option C: (P) → (3), (Q) → (4), (R) → (1), (S) → (5).

The given point is $$P(1,3,2)$$ and the line parallel to which we must draw a line through $$P$$ is
$$\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}$$

The direction ratios of this line are $$1,2,1$$, so a line through $$P$$ with the same direction has parametric form
$$x = 1 + t,\; y = 3 + 2t,\; z = 2 + t \qquad -(1)$$

This line meets the plane $$L_1 : x - y + 3z = 6$$ at the point $$Q$$. Substitute the coordinates from $$(1)$$ into the plane:

$$\bigl(1+t\bigr) - \bigl(3+2t\bigr) + 3\bigl(2+t\bigr) = 6$$
$$1+t - 3 - 2t + 6 + 3t = 6$$
$$4 + 2t = 6 \;\;\Longrightarrow\;\; t = 1$$

Putting $$t=1$$ back in $$(1)$$ gives
$$Q(2,\,5,\,3)$$

Vector $$\overrightarrow{PQ} = (2-1,\,5-3,\,3-2) = (1,2,1)$$, so
$$PQ = \sqrt{1^{2}+2^{2}+1^{2}} = \sqrt{6}$$
Thus statement A is true.

A line through $$Q$$ perpendicular to plane $$L_1$$ must have direction equal to the normal of $$L_1$$.
Since $$L_1$$ is $$x - y + 3z = 6$$, its normal vector is $$(1,-1,3)$$.

Parametric equations of this perpendicular line are
$$x = 2 + s,\; y = 5 - s,\; z = 3 + 3s \qquad -(2)$$

This line meets the plane $$L_2 : 2x - y + z = -4$$ at point $$R$$. Insert the coordinates from $$(2)$$ into $$L_2$$:

$$2(2+s) - (5 - s) + (3 + 3s) = -4$$
$$4 + 2s - 5 + s + 3 + 3s = -4$$
$$2 + 6s = -4$$
$$6s = -6 \;\;\Longrightarrow\;\; s = -1$$

Substituting $$s=-1$$ in $$(2)$$ gives
$$R(1,\,6,\,0)$$

Thus statement B, which claimed $$R(1,6,3)$$, is false.

Centroid $$G$$ of $$\triangle PQR$$ is
$$G\bigl(\,\tfrac{1+2+1}{3},\;\tfrac{3+5+6}{3},\;\tfrac{2+3+0}{3}\bigr) = \Bigl(\tfrac{4}{3},\;\tfrac{14}{3},\;\tfrac{5}{3}\Bigr)$$

Hence statement C is true.

For statement D, compute the remaining two sides:

$$QR = \sqrt{(1)^{2} + (1)^{2} + (-3)^{2}} = \sqrt{11}$$
$$RP = \sqrt{(0)^{2} + (-3)^{2} + 2^{2}} = \sqrt{13}$$

Perimeter $$= PQ + QR + RP = \sqrt{6} + \sqrt{11} + \sqrt{13}$$, not $$\sqrt{2} + \sqrt{6} + \sqrt{11}$$, so statement D is false.

Therefore the correct statements are:
Option A and Option C.

The formula for the image of point $$(x_1, y_1)$$ in the line $$ax + by + c = 0$$ is:

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$

Here the line is $$x + 2y - 2 = 0$$, so $$a = 1, b = 2, c = -2$$, and $$(x_1, y_1) = (-4, 5)$$.

$$\frac{x - (-4)}{1} = \frac{y - 5}{2} = \frac{-2((-4) + 2(5) - 2)}{1 + 4} = \frac{-2(4)}{5} = \frac{-8}{5}$$

$$x + 4 = -\frac{8}{5} \Rightarrow x = -4 - \frac{8}{5} = -\frac{28}{5}$$

$$y - 5 = -\frac{16}{5} \Rightarrow y = 5 - \frac{16}{5} = \frac{9}{5}$$

The image is $$\left(-\frac{28}{5}, \frac{9}{5}\right)$$.

Now check if this lies on $$(x+4)^2 + (y-3)^2 = r^2$$:

$$\left(-\frac{28}{5}+4\right)^2 + \left(\frac{9}{5}-3\right)^2 = \left(-\frac{8}{5}\right)^2 + \left(-\frac{6}{5}\right)^2 = \frac{64}{25} + \frac{36}{25} = \frac{100}{25} = 4$$

So $$r^2 = 4$$, which gives $$r = 2$$.

For infinitely many solutions, the third equation must be a linear combination of the first two.

Let the third equation = $$\alpha$$(first) + $$\beta$$(second):

$$3\alpha + 7\beta = 97$$ ... (i), $$5\alpha + 11\beta = 155$$ ... (ii), $$\lambda\alpha - 9\beta = -189$$ ... (iii), $$3\alpha + 2\beta = \mu$$ ... (iv)

From (i) and (ii): subtracting gives $$2\beta = 20$$, so $$\beta = 10$$ and $$\alpha = 9$$.

From (iii): $$9\lambda - 90 = -189$$, so $$\lambda = -11$$.

From (iv): $$\mu = 3(9) + 2(10) = 47$$.

$$\mu + 2\lambda = 47 + 2(-11) = 25$$.

The correct answer is Option 2: 25.

Write $$X=(x,y,z)$$. Using the distance formula,

$$(dist(X,P))^{2}- (dist(X,Q))^{2}$$
$$=(x-1)^{2}+(y-2)^{2}+(z-3)^{2}-\left[(x-4)^{2}+(y-2)^{2}+(z-7)^{2}\right]$$
$$=\left[(x-1)^{2}-(x-4)^{2}\right]+\left[(z-3)^{2}-(z-7)^{2}\right]$$
$$=(6x-15)+(8z-40)=6x+8z-55.$$ Therefore

Set $$S$$ : $$6x+8z-55=50\;\Longrightarrow\;6x+8z=105$$.

Set $$T$$ : $$-(6x+8z-55)=50\;\Longrightarrow\;6x+8z=5$$.

Hence $$S$$ and $$T$$ are two parallel planes perpendicular to the vector $$(6,0,8)$$. Their separation is

$$\frac{|105-5|}{\sqrt{6^{2}+0^{2}+8^{2}}}=\frac{100}{10}=10.$$

Checking the options

Option A  In the infinite plane $$6x+8z=105$$ we can choose any three non-collinear points. Scale them so that the area becomes exactly $$1$$ (e.g. first pick a triangle of area $$k$$ and then take all its vertices closer to their centroid by a factor $$\sqrt{k}$$). Thus a triangle of area $$1$$ with all vertices in $$S$$ exists. Option A is true.

Option B  Because $$T$$ is a plane, pick any two distinct points $$L,M\in T$$; the entire segment $$LM$$ lies in the same plane and therefore in $$T$$. Option B is true.

Option C  Let $$\mathbf n=(6,0,8)$$ and $$\|\mathbf n\|=10$$. For any point $$A\in S$$ define $$D=A-\mathbf n$$ (so $$D\in T$$). Choose a vector $$\mathbf v$$ lying in the plane (orthogonal to $$\mathbf n$$) with length $$14$$. Put $$B=A+\mathbf v$$ and $$C=D+\mathbf v$$. Then $$AB=CD=14$$ lie in $$S$$, $$BC=DA=10$$ join the two planes, and the quadrilateral $$ABCD$$ is a rectangle of perimeter $$2(14+10)=48$$ having two vertices in $$S$$ and two in $$T$$. Because we may pick $$A$$ anywhere in $$S$$ and rotate $$\mathbf v$$ freely in the plane, infinitely many such rectangles exist. Option C is true.

Option D  A square of perimeter $$48$$ has side $$12$$. If two consecutive vertices are in different planes, every side must join the planes, so each side’s normal component equals the plane distance $$10$$. Hence the in-plane component of a side has length $$\sqrt{12^{2}-10^{2}}=2\sqrt{11}$$. Take two adjacent sides $$\mathbf a,\mathbf b$$; their normal components are $$\pm10$$ along the same normal, while their in-plane parts are orthogonal. The dot product condition for a square demands $$\mathbf a\cdot\mathbf b=0$$, that is
$$-10^{2} + (\text{dot product of in-plane parts}) = 0 \;\Longrightarrow\; (\text{dot product of in-plane parts}) = 100.$$
But the two in-plane vectors each have magnitude $$2\sqrt{11}$$, so their dot product is at most $$(2\sqrt{11})^{2}=44\lt100.$$ Therefore such a square cannot be formed. Option D is false.

The true statements are: Option A, Option B, and Option C.

Option A, Option B, Option C

For infinitely many solutions, the system must satisfy:

1. Determinant = 0
2. Equations are consistent (one equation is a linear combination of others)

Determinant condition gives:
$$\mu=0$$

Now enforce consistency. From coefficient comparison:
$$R_3=-2R_1+R_2$$

Apply same to constants:
$$-1=-2\lambda+(-3)$$
$$\Rightarrow-2\lambda-3=-1$$
$$\Rightarrow\lambda=-1$$

Finally:
$$2\mu+3\lambda=2(0)+3(-1)=-3$$

The system of equations is:

$$2x + 3y - z = 5 \quad \text{(1)}$$

$$x + \alpha y + 3z = -4 \quad \text{(2)}$$

$$3x - y + \beta z = 7 \quad \text{(3)}$$

For the system to have infinitely many solutions, the equations must be linearly dependent. This means the third equation can be expressed as a linear combination of the first two. Assume:

$$\text{Eq. (3)} = p \cdot \text{Eq. (1)} + q \cdot \text{Eq. (2)}$$

Equating coefficients for each variable and the constant term:

For $$x$$: $$3 = 2p + q \quad \text{(4)}$$

For $$y$$: $$-1 = 3p + \alpha q \quad \text{(5)}$$

For $$z$$: $$\beta = -p + 3q \quad \text{(6)}$$

For constants: $$7 = 5p - 4q \quad \text{(7)}$$

Solve equations (4) and (7) for $$p$$ and $$q$$. From equation (4):

$$q = 3 - 2p$$

Substitute into equation (7):

$$5p - 4(3 - 2p) = 7$$

$$5p - 12 + 8p = 7$$

$$13p = 19$$

$$p = \frac{19}{13}$$

Then:

$$q = 3 - 2 \left( \frac{19}{13} \right) = \frac{39}{13} - \frac{38}{13} = \frac{1}{13}$$

Substitute $$p$$ and $$q$$ into equation (5) to find $$\alpha$$:

$$3 \left( \frac{19}{13} \right) + \alpha \left( \frac{1}{13} \right) = -1$$

$$\frac{57}{13} + \frac{\alpha}{13} = -1$$

Multiply both sides by 13:

$$57 + \alpha = -13$$

$$\alpha = -70$$

Substitute $$p$$ and $$q$$ into equation (6) to find $$\beta$$:

$$\beta = -\left( \frac{19}{13} \right) + 3 \left( \frac{1}{13} \right) = \frac{-19 + 3}{13} = \frac{-16}{13}$$

Now compute $$13\alpha\beta$$:

$$13 \alpha \beta = 13 \times (-70) \times \left( \frac{-16}{13} \right) = 13 \times \frac{1120}{13} = 1120$$

Thus, $$13\alpha\beta = 1120$$.

Verification: The determinant of the coefficient matrix is zero, and the linear dependence condition is satisfied, confirming infinitely many solutions.

The value 1120 corresponds to option B.

We need to find $$\lambda + 2\mu$$ such that the system $$x + 2y + 3z = 5$$, $$2x + 3y + z = 9$$, $$4x + 3y + \lambda z = \mu$$ has infinitely many solutions.

For a system of 3 equations in 3 unknowns to have infinitely many solutions, the third equation must be a linear combination of the first two, AND the determinant of the coefficient matrix must be zero.

Let the third equation = $$\alpha$$ (Eq 1) + $$\beta$$ (Eq 2):

$$\alpha(x + 2y + 3z) + \beta(2x + 3y + z) = \alpha \cdot 5 + \beta \cdot 9$$

$$(\alpha + 2\beta)x + (2\alpha + 3\beta)y + (3\alpha + \beta)z = 5\alpha + 9\beta$$

Comparing with $$4x + 3y + \lambda z = \mu$$:

$$\alpha + 2\beta = 4$$ ... (i)

$$2\alpha + 3\beta = 3$$ ... (ii)

$$3\alpha + \beta = \lambda$$ ... (iii)

$$5\alpha + 9\beta = \mu$$ ... (iv)

From (i): $$\alpha = 4 - 2\beta$$.

Substitute into (ii): $$2(4 - 2\beta) + 3\beta = 3 \implies 8 - 4\beta + 3\beta = 3 \implies \beta = 5$$.

So $$\alpha = 4 - 10 = -6$$.

$$\lambda = 3(-6) + 5 = -18 + 5 = -13$$

$$\mu = 5(-6) + 9(5) = -30 + 45 = 15$$

$$\lambda + 2\mu = -13 + 2(15) = -13 + 30 = 17$$

The correct answer is 17 (Option B).

We have $$A(1, 3, 2)$$, $$B(-2, 8, 0)$$, $$C(3, 6, 7)$$.

$$\vec{AB} = (-3, 5, -2)$$, so $$AB = \sqrt{9 + 25 + 4} = \sqrt{38}$$

$$\vec{AC} = (2, 3, 5)$$, so $$AC = \sqrt{4 + 9 + 25} = \sqrt{38}$$

Since $$AB = AC$$, the triangle is isosceles. By the angle bisector theorem, D divides BC in the ratio $$AB : AC = 1 : 1$$, so D is the midpoint of BC.

$$D = \left(\frac{-2+3}{2}, \frac{8+6}{2}, \frac{0+7}{2}\right) = \left(\frac{1}{2}, 7, \frac{7}{2}\right)$$

$$\vec{AD} = \left(\frac{1}{2} - 1, 7 - 3, \frac{7}{2} - 2\right) = \left(-\frac{1}{2}, 4, \frac{3}{2}\right)$$

The projection of $$\vec{AD}$$ on $$\vec{AC}$$ is given by:

$$\text{Projection} = \frac{\vec{AD} \cdot \vec{AC}}{|\vec{AC}|}$$

$$\vec{AD} \cdot \vec{AC} = \left(-\frac{1}{2}\right)(2) + (4)(3) + \left(\frac{3}{2}\right)(5) = -1 + 12 + \frac{15}{2} = 11 + \frac{15}{2} = \frac{37}{2}$$

$$|\vec{AC}| = \sqrt{38}$$

$$\text{Projection} = \frac{37/2}{\sqrt{38}} = \frac{37}{2\sqrt{38}}$$

$$A(2,-3,3)$$, $$B(2,2,3)$$, $$C(-1,1,3)$$.

$$\vec{AB} = (0,5,0)$$, $$AB = 5$$.

$$\vec{AC} = (-3,4,0)$$, $$AC = 5$$.

Since $$AB = AC$$, D is the midpoint of BC (angle bisector theorem with equal sides).

$$D = \frac{B + C}{2} = \left(\frac{1}{2}, \frac{3}{2}, 3\right)$$

$$\vec{AD} = \left(\frac{1}{2} - 2, \frac{3}{2}+3, 0\right) = \left(-\frac{3}{2}, \frac{9}{2}, 0\right)$$

$$l^2 = \frac{9}{4} + \frac{81}{4} = \frac{90}{4} = \frac{45}{2}$$

$$2l^2 = 45$$.

The answer is $$45$$, which corresponds to Option (4).

For an obtuse angle, the dot product $$\vec{a} \cdot \vec{b} < 0$$.

$$(\alpha t)(t) + (6)(-2) + (-3)(-2\alpha t) < 0$$

$$\alpha t^2 + 6\alpha t - 12 < 0$$.

For a quadratic $$At^2 + Bt + C$$ to be always negative:

$$A < 0 \implies \mathbf{\alpha < 0}$$.

Discriminant $$D < 0 \implies (6\alpha)^2 - 4(\alpha)(-12) < 0$$.

$$36\alpha^2 + 48\alpha < 0 \implies 12\alpha(3\alpha + 4) < 0$$.

The roots are $$0$$ and $$-4/3$$. The inequality holds for $$\alpha \in (-4/3, 0)$$.

Checking $$\alpha = 0$$: The equation becomes $$-12 < 0$$, which is true. So, $$\alpha \in \mathbf{(-4/3, 0]}$$. (Option C)

Standardize the direction ratios (DRs):

$$L_1: \frac{x-2}{-3} = \frac{y-2/3}{(4\lambda+1)/3} = \frac{z-4}{-1}$$. DRs: $$\vec{v_1} = (-3, \frac{4\lambda+1}{3}, -1)$$.

$$L_2: \frac{x+3}{3\mu} = \frac{y-1/2}{-3} = \frac{z-5}{-7}$$. DRs: $$\vec{v_2} = (3\mu, -3, -7)$$.

For perpendicular lines, $$\vec{v_1} \cdot \vec{v_2} = 0$$:

$$(-3)(3\mu) + (\frac{4\lambda+1}{3})(-3) + (-1)(-7) = 0$$

$$-9\mu - (4\lambda + 1) + 7 = 0$$

$$-9\mu - 4\lambda + 6 = 0 \implies \mathbf{4\lambda + 9\mu = 6}$$

We need to find the mirror image of $$P(3, 4, 9)$$ in the line $$\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$$.

Since we need the foot of the perpendicular from P to the line, we take a point on the line, $$A(1, -1, 2)$$, with direction vector $$\vec{d} = (3, 2, 1)$$.

A general point on the line is $$(1 + 3t,\,-1 + 2t,\,2 + t)$$, so the vector from this point to P is $$(3 - (1 + 3t),\,4 - (-1 + 2t),\,9 - (2 + t)) = (2 - 3t,\,5 - 2t,\,7 - t)\,.$$

For the foot of perpendicular, this vector must be perpendicular to $$\vec{d}$$, so

$$ (2 - 3t)(3) + (5 - 2t)(2) + (7 - t)(1) = 0 $$

$$ 6 - 9t + 10 - 4t + 7 - t = 0 $$

$$ 23 - 14t = 0 \implies t = \frac{23}{14}\,.$$

Substituting back gives

$$ N = \Bigl(1 + 3\cdot\frac{23}{14},\,-1 + 2\cdot\frac{23}{14},\,2 + \frac{23}{14}\Bigr) = \Bigl(\frac{83}{14},\,\frac{32}{14},\,\frac{51}{14}\Bigr) = \Bigl(\frac{83}{14},\,\frac{16}{7},\,\frac{51}{14}\Bigr)\,.$$

Now, since N is the midpoint of P and its mirror image Q, we set

$$ \alpha = 2\cdot\frac{83}{14} - 3 = \frac{166}{14} - \frac{42}{14} = \frac{124}{14} = \frac{62}{7}, $$

$$ \beta = 2\cdot\frac{16}{7} - 4 = \frac{32}{7} - \frac{28}{7} = \frac{4}{7}, $$

$$ \gamma = 2\cdot\frac{51}{14} - 9 = \frac{102}{14} - \frac{126}{14} = -\frac{24}{14} = -\frac{12}{7}. $$

Therefore,

$$ \alpha + \beta + \gamma = \frac{62}{7} + \frac{4}{7} - \frac{12}{7} = \frac{54}{7}, $$

and

$$ 14(\alpha + \beta + \gamma) = 14 \times \frac{54}{7} = 2 \times 54 = 108. $$

The correct answer is Option 3: 108.

The shortest distance between two skew lines $$L_1: \vec{r} = \vec{a} + \lambda \vec{b}$$ and $$L_2: \vec{r} = \vec{c} + \mu \vec{d}$$ is given by the formula:

$$d = \frac{ | (\vec{b} \times \vec{d}) \cdot (\vec{a} - \vec{c}) | }{ |\vec{b} \times \vec{d}| }$$

First, rewrite the given lines in standard form.

For $$L_1: \vec{r} = (2 + \lambda)\hat{i} + (1 - 3\lambda)\hat{j} + (3 + 4\lambda)\hat{k}$$:

Position vector $$\vec{a} = 2\hat{i} + 1\hat{j} + 3\hat{k}$$

Direction vector $$\vec{b} = 1\hat{i} - 3\hat{j} + 4\hat{k}$$

For $$L_2: \vec{r} = 2(1 + \mu)\hat{i} + 3(1 + \mu)\hat{j} + (5 + \mu)\hat{k} = (2 + 2\mu)\hat{i} + (3 + 3\mu)\hat{j} + (5 + \mu)\hat{k}$$:

Position vector $$\vec{c} = 2\hat{i} + 3\hat{j} + 5\hat{k}$$

Direction vector $$\vec{d} = 2\hat{i} + 3\hat{j} + 1\hat{k}$$

Now, compute $$\vec{a} - \vec{c}$$:

$$\vec{a} - \vec{c} = (2 - 2)\hat{i} + (1 - 3)\hat{j} + (3 - 5)\hat{k} = 0\hat{i} - 2\hat{j} - 2\hat{k}$$

Next, compute the cross product $$\vec{b} \times \vec{d}$$:

$$\vec{b} = \begin{pmatrix} 1 \\ -3 \\ 4 \end{pmatrix}, \vec{d} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$$

$$\vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i} \left[ (-3)(1) - (4)(3) \right] - \hat{j} \left[ (1)(1) - (4)(2) \right] + \hat{k} \left[ (1)(3) - (-3)(2) \right]$$

Calculate each component:

i-component: $$(-3)(1) - (4)(3) = -3 - 12 = -15$$

j-component: $$- \left[ (1)(1) - (4)(2) \right] = - \left[ 1 - 8 \right] = -(-7) = 7$$

k-component: $$(1)(3) - (-3)(2) = 3 - (-6) = 3 + 6 = 9$$

Thus, $$\vec{b} \times \vec{d} = -15\hat{i} + 7\hat{j} + 9\hat{k}$$

Now, find the magnitude $$|\vec{b} \times \vec{d}|$$:

$$|\vec{b} \times \vec{d}| = \sqrt{(-15)^2 + (7)^2 + (9)^2} = \sqrt{225 + 49 + 81} = \sqrt{355}$$

Compute the scalar triple product $$(\vec{b} \times \vec{d}) \cdot (\vec{a} - \vec{c})$$:

$$\vec{b} \times \vec{d} = -15\hat{i} + 7\hat{j} + 9\hat{k}, \quad \vec{a} - \vec{c} = 0\hat{i} - 2\hat{j} - 2\hat{k}$$

Dot product: $$(-15)(0) + (7)(-2) + (9)(-2) = 0 - 14 - 18 = -32$$

Absolute value: $$| -32 | = 32$$

Therefore, the shortest distance is:

$$d = \frac{32}{\sqrt{355}}$$

The distance is given as $$\frac{m}{\sqrt{n}}$$ with $$\gcd(m, n) = 1$$. Here, $$m = 32$$ and $$n = 355$$.

Check $$\gcd(32, 355)$$:

$$32 = 2^5, \quad 355 = 5 \times 71$$

Since there are no common prime factors, $$\gcd(32, 355) = 1$$.

Thus, $$m = 32$$, $$n = 355$$, and $$m + n = 32 + 355 = 387$$.

Verify that the lines are skew (neither parallel nor intersecting):

Direction vectors $$\vec{b} = \langle 1, -3, 4 \rangle$$, $$\vec{d} = \langle 2, 3, 1 \rangle$$.

Check for parallelism: Is there $$k$$ such that $$\langle 1, -3, 4 \rangle = k \langle 2, 3, 1 \rangle$$?

From $$1 = 2k$$, $$k = \frac{1}{2}$$, but $$-3 = 3 \times \frac{1}{2} = 1.5 \neq -3$$, so not parallel.

Check for intersection: Set equations equal:

i-component: $$2 + \lambda = 2 + 2\mu \implies \lambda = 2\mu$$

j-component: $$1 - 3\lambda = 3 + 3\mu$$

Substitute $$\lambda = 2\mu$$: $$1 - 3(2\mu) = 3 + 3\mu \implies 1 - 6\mu = 3 + 3\mu \implies -6\mu - 3\mu = 3 - 1 \implies -9\mu = 2 \implies \mu = -\frac{2}{9}$$

Then $$\lambda = 2 \times -\frac{2}{9} = -\frac{4}{9}$$

k-component: Left side: $$3 + 4\lambda = 3 + 4(-\frac{4}{9}) = 3 - \frac{16}{9} = \frac{27}{9} - \frac{16}{9} = \frac{11}{9}$$

Right side: $$5 + \mu = 5 - \frac{2}{9} = \frac{45}{9} - \frac{2}{9} = \frac{43}{9}$$

$$\frac{11}{9} \neq \frac{43}{9}$$, so no intersection. Thus, lines are skew.

The value of $$m + n$$ is 387, which corresponds to option D.

line is perpendicular to both lines whose direction vectors are $$\vec{d_1} = (2, 3, 5)$$ and $$\vec{d_2} = (-1, 3, 2)$$, its direction is given by the cross product:

$$\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix}$$

The component along $$\hat{i}$$ is $$(3)(2) - (5)(3) = 6 - 15 = -9,$$ along $$\hat{j}$$ is $$-((2)(2) - (5)(-1)) = -(4+5) = -9,$$ and along $$\hat{k}$$ is $$(2)(3) - (3)(-1) = 6+3 = 9.$$ Thus $$\vec{d} = (-9, -9, 9),$$ which simplifies to $$(1, 1, -1).$$

A line through $$P(5,-4,3)$$ with direction $$(1,1,-1)$$ can be parametrised as $$(x, y, z) = (5+t, -4+t, 3-t).$$ 

The vector from a general point on this line to $$Q$$ is $$\vec{PQ_t} = (0-(5+t), 2-(-4+t), -2-(3-t)) = (-5-t, 6-t, -5+t).$$ For the foot of the perpendicular, this vector must be perpendicular to $$\vec{d} = (1,1,-1),$$ so we set

$$(-5-t)(1) + (6-t)(1) + (-5+t)(-1) = 0$$

$$6-3t = 0 \implies t = 2.$$

At $$t = 2$$, $$(5+2, -4+2, 3-2) = (7, -2, 1)$$ gives the foot of the perpendicular.

$$d = \sqrt{(0-7)^2 + (2-(-2))^2 + (-2-1)^2} = \sqrt{49 + 16 + 9} = \sqrt{74}$$

Line $$L$$ passes through $$(1,2,3)$$ and $$(2,3,5)$$, and the problem asks for the distance of $$P\left(\frac{11}{3},\frac{11}{3},\frac{19}{3}\right)$$ from $$L$$ along the line $$\frac{3x-11}{2}=\frac{3y-11}{1}=\frac{3z-19}{2}$$.

Since $$L$$ passes through these two points, its direction vector is $$(2-1,3-2,5-3)=(1,1,2)$$, so that $$L:\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda$$.

Next, the line $$\frac{3x-11}{2}=\frac{3y-11}{1}=\frac{3z-19}{2}$$ passes through $$P\left(\frac{11}{3},\frac{11}{3},\frac{19}{3}\right)$$ with direction vector $$(2,1,2)$$, which leads to the parametric form $$\left(\frac{11}{3}+\frac{2t}{3},\frac{11}{3}+\frac{t}{3},\frac{19}{3}+\frac{2t}{3}\right)$$.

For this point to lie on $$L$$, the relations $$\frac{\left(\frac{11}{3}+\frac{2t}{3}\right)-1}{1}=\frac{\left(\frac{11}{3}+\frac{t}{3}\right)-2}{1}$$ yield $$\frac{8+2t}{3}=\frac{5+t}{3}$$, which leads to $$8+2t=5+t$$, so that $$t=-3$$.

Substituting $$t=-3$$ gives the intersection point $$\left(\frac{11}{3}-2,\frac{11}{3}-1,\frac{19}{3}-2\right)=\left(\frac{5}{3},\frac{8}{3},\frac{13}{3}\right)$$, and verifying on $$L$$ shows $$\frac{5/3-1}{1}=\frac{8/3-2}{1}=\frac{13/3-3}{2}=\frac{2}{3}$$.

Then the displacement from $$P$$ to the intersection point corresponds to $$t=-3$$ along the direction vector $$\left(\frac{2}{3},\frac{1}{3},\frac{2}{3}\right)$$, so $$d=\sqrt{\left(\frac{2(-3)}{3}\right)^2+\left(\frac{-3}{3}\right)^2+\left(\frac{2(-3)}{3}\right)^2}=\sqrt{4+1+4}=\sqrt{9}=3$$.

The distance is 3.

The correct answer is Option (4): 3.

The shortest distance between two skew lines is given by the formula:

$$d = \left| \frac{ (\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) }{ |\vec{b_1} \times \vec{b_2}| } \right|$$

where $$\vec{a_1}$$ and $$\vec{a_2}$$ are points on the lines, and $$\vec{b_1}$$ and $$\vec{b_2}$$ are the direction vectors.

For the first line: $$\frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}$$, a point is $$\vec{a_1} = \langle \lambda, 2, 1 \rangle$$ and direction vector $$\vec{b_1} = \langle -2, 1, 1 \rangle$$.

For the second line: $$\frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}$$, a point is $$\vec{a_2} = \langle \sqrt{3}, 1, 2 \rangle$$ and direction vector $$\vec{b_2} = \langle 1, -2, 1 \rangle$$.

The vector between the points is $$\vec{a_2} - \vec{a_1} = \langle \sqrt{3} - \lambda, -1, 1 \rangle$$.

Compute the cross product $$\vec{b_1} \times \vec{b_2}$$:

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 1 \cdot (-2)) - \hat{j}(-2 \cdot 1 - 1 \cdot 1) + \hat{k}(-2 \cdot (-2) - 1 \cdot 1)$$

$$= \hat{i}(1 + 2) - \hat{j}(-2 - 1) + \hat{k}(4 - 1) = \hat{i}(3) - \hat{j}(-3) + \hat{k}(3) = \langle 3, 3, 3 \rangle$$

The magnitude is $$|\vec{b_1} \times \vec{b_2}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$$.

The scalar triple product is:

$$(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) = \langle 3, 3, 3 \rangle \cdot \langle \sqrt{3} - \lambda, -1, 1 \rangle = 3(\sqrt{3} - \lambda) + 3(-1) + 3(1) = 3\sqrt{3} - 3\lambda - 3 + 3 = 3\sqrt{3} - 3\lambda$$

The distance is:

$$d = \left| \frac{3\sqrt{3} - 3\lambda}{3\sqrt{3}} \right| = \left| \frac{\sqrt{3} - \lambda}{\sqrt{3}} \right| = \left| 1 - \frac{\lambda}{\sqrt{3}} \right|$$

Given that $$d = 1$$:

$$\left| 1 - \frac{\lambda}{\sqrt{3}} \right| = 1$$

This absolute value equation gives two cases:

Case 1: $$1 - \frac{\lambda}{\sqrt{3}} = 1$$

$$-\frac{\lambda}{\sqrt{3}} = 0 \implies \lambda = 0$$

Case 2: $$1 - \frac{\lambda}{\sqrt{3}} = -1$$

$$-\frac{\lambda}{\sqrt{3}} = -2 \implies \frac{\lambda}{\sqrt{3}} = 2 \implies \lambda = 2\sqrt{3}$$

The possible values of $$\lambda$$ are $$0$$ and $$2\sqrt{3}$$. The sum is $$0 + 2\sqrt{3} = 2\sqrt{3}$$.

Verification:

For $$\lambda = 0$$, the numerator is $$3\sqrt{3} - 3(0) = 3\sqrt{3}$$, so $$d = \left| \frac{3\sqrt{3}}{3\sqrt{3}} \right| = 1$$.

For $$\lambda = 2\sqrt{3}$$, the numerator is $$3\sqrt{3} - 3(2\sqrt{3}) = 3\sqrt{3} - 6\sqrt{3} = -3\sqrt{3}$$, so $$d = \left| \frac{-3\sqrt{3}}{3\sqrt{3}} \right| = 1$$.

Thus, the sum of all possible values of $$\lambda$$ is $$2\sqrt{3}$$.

Consider the two lines given by

$$L_1: \frac{x-\lambda}{2} = \frac{y-4}{3} = \frac{z-3}{4}$$ (direction $$\vec{d_1} = (2,3,4)$$, point $$A = (\lambda,4,3)$$)

$$L_2: \frac{x-2}{4} = \frac{y-4}{6} = \frac{z-7}{8}$$ (direction $$\vec{d_2} = (4,6,8)$$, point $$B = (2,4,7)$$)

Since $$\vec{d_2} = 2\vec{d_1}$$, the two lines are parallel. For parallel lines, the shortest distance is given by

$$d = \frac{|\vec{AB} \times \vec{d_1}|}{|\vec{d_1}|}$$

Here, $$\vec{AB} = B - A = (2-\lambda, 0, 4)$$ and

$$\vec{AB} \times \vec{d_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2-\lambda & 0 & 4 \\ 2 & 3 & 4 \end{vmatrix}$$

$$= \hat{i}(0 \cdot 4 - 4 \cdot 3) - \hat{j}((2-\lambda) \cdot 4 - 4 \cdot 2) + \hat{k}((2-\lambda) \cdot 3 - 0 \cdot 2)$$

$$= \hat{i}(-12) - \hat{j}(8-4\lambda-8) + \hat{k}(6-3\lambda)$$

$$= -12\hat{i} + 4\lambda\hat{j} + (6-3\lambda)\hat{k}$$

It follows that

$$|\vec{AB} \times \vec{d_1}| = \sqrt{144 + 16\lambda^2 + (6-3\lambda)^2} = \sqrt{144 + 16\lambda^2 + 36 - 36\lambda + 9\lambda^2}$$

$$= \sqrt{25\lambda^2 - 36\lambda + 180}$$

and

$$|\vec{d_1}| = \sqrt{4+9+16} = \sqrt{29} \,.$$

Thus the distance is

$$d = \frac{\sqrt{25\lambda^2 - 36\lambda + 180}}{\sqrt{29}} = \frac{13}{\sqrt{29}}$$

Squaring both sides gives

$$25\lambda^2 - 36\lambda + 180 = 169$$

$$25\lambda^2 - 36\lambda + 11 = 0$$

Using the quadratic formula,

$$\lambda = \frac{36 \pm \sqrt{1296 - 1100}}{50} = \frac{36 \pm \sqrt{196}}{50} = \frac{36 \pm 14}{50}$$

$$\lambda = \frac{50}{50} = 1 \quad \text{or} \quad \lambda = \frac{22}{50} = \frac{11}{25}$$

From the options, $$\lambda = 1$$ is available.

The correct answer is Option 4: 1.

The line is $$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = t$$ and a general point on the line is $$(5t-3,\,2t+1,\,3t-4)\,.$$

The foot of the perpendicular from $$(1,2,3)$$ to this line satisfies $$\vec{PF}\cdot\vec{d}=0$$ where $$\vec{d}=(5,2,3)\,.$$

Substituting the coordinates into the dot product gives $$ (5t-3-1)(5) + (2t+1-2)(2) + (3t-4-3)(3) = 0 $$ which simplifies to $$5(5t-4) + 2(2t-1) + 3(3t-7) = 0$$ and to $$25t - 20 + 4t - 2 + 9t - 21 = 0$$ yielding $$38t - 43 = 0 \Rightarrow t = \frac{43}{38}\,.$$

So the foot of the perpendicular $$(\alpha,\beta,\gamma)$$ has coordinates $$\alpha = 5\cdot\frac{43}{38} - 3 = \frac{215 - 114}{38} = \frac{101}{38},\quad \beta = 2\cdot\frac{43}{38} + 1 = \frac{86 + 38}{38} = \frac{124}{38} = \frac{62}{19},\quad \gamma = 3\cdot\frac{43}{38} - 4 = \frac{129 - 152}{38} = \frac{-23}{38}\,.$$

Then $$\alpha+\beta+\gamma = \frac{101}{38} + \frac{124}{38} + \frac{-23}{38} = \frac{202}{38} = \frac{101}{19},$$ and $$19(\alpha+\beta+\gamma) = 19\times\frac{101}{19} = 101\,.$$

The answer is Option (2): $$\boxed{101}$$.

Find the image of point $$(8, 5, 7)$$ in the line $$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}$$.

The line passes through $$A(1, -1, 2)$$ and has direction vector $$\vec{d} = (2, 3, 5)$$. Let the foot of the perpendicular from $$P(8, 5, 7)$$ to the line be $$F = (1+2t,\,-1+3t,\;2+5t)$$, so that $$\vec{PF} = (2t-7,\;3t-6,\;5t-5)$$.

Enforcing $$\vec{PF}\cdot\vec{d}=0$$ yields $$ 2(2t-7) + 3(3t-6) + 5(5t-5) = 0 $$ $$ 4t - 14 + 9t - 18 + 25t - 25 = 0 $$ $$ 38t - 57 = 0 $$ and hence $$ t = \frac{3}{2} $$.

Substituting back gives $$F = (1+3,\,-1+4.5,\;2+7.5) = (4,\;3.5,\;9.5)$$. Since $$F$$ is the midpoint of $$P$$ and its image $$(\alpha,\beta,\gamma)$$, we get $$ \alpha = 2(4) - 8 = 0, \quad \beta = 2(3.5) - 5 = 2, \quad \gamma = 2(9.5) - 7 = 12. $$

Therefore, $$ \alpha + \beta + \gamma = 0 + 2 + 12 = 14 $$. The correct answer is Option (3): 14.

Line 1: $$\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1} = s$$. Point: $$(-6+3s, 2s, -1+s)$$.

Line 2: $$\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2} = t$$. Point: $$(7+4t, 9+3t, 4+2t)$$.

At intersection:

From (3): $$s = 5 + 2t$$. Substitute in (2): $$2(5+2t) - 3t = 9 \Rightarrow 10 + t = 9 \Rightarrow t = -1$$.

$$s = 5 - 2 = 3$$.

Check (1): $$9 - (-4) = 13$$ ✓.

Intersection point: $$(-6+9, 6, -1+3) = (3, 6, 2)$$.

Distance from $$(3,6,2)$$ to $$(7,8,9)$$:

The correct answer is Option (4): 75.

$$L_1$$ direction: $$(1, -1, 2)$$. $$L_2$$ direction: $$(3, 1, p)$$.

$$L_1 \perp L_2$$: $$3 - 1 + 2p = 0 \Rightarrow p = -1$$. $$L_2$$ direction: $$(3, 1, -1)$$.

$$L_3$$ direction $$(l, m, n)$$ perpendicular to both $$L_1$$ and $$L_2$$:

$$(l, m, n) = (1, -1, 2) \times (3, 1, -1) = (1-2, 6+1, 1+3) = (-1, 7, 4)$$.

$$L_3$$: $$\vec{r} = \delta(-\hat{i} + 7\hat{j} + 4\hat{k})$$. Points on $$L_3$$: $$(-\delta, 7\delta, 4\delta)$$.

For $$\delta = 1$$: $$(-1, 7, 4)$$.

The answer is Option (1): $$\boxed{(-1, 7, 4)}$$.

The line is $$\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2} = t$$.

A general point on the line is $$(8t - 3, 2t + 4, 2t - 1)$$.

The distance from this point to $$R(1, 2, 3)$$ is 6:

$$(8t - 4)^2 + (2t + 2)^2 + (2t - 4)^2 = 36$$

$$64t^2 - 64t + 16 + 4t^2 + 8t + 4 + 4t^2 - 16t + 16 = 36$$

$$72t^2 - 72t + 36 = 36$$

$$72t^2 - 72t = 0$$

$$72t(t - 1) = 0$$

So $$t = 0$$ or $$t = 1$$.

For $$t = 0$$: $$P = (-3, 4, -1)$$

For $$t = 1$$: $$Q = (5, 6, 1)$$

The centroid of triangle PQR:

$$\alpha = \frac{-3 + 5 + 1}{3} = 1$$

$$\beta = \frac{4 + 6 + 2}{3} = 4$$

$$\gamma = \frac{-1 + 1 + 3}{3} = 1$$

$$\alpha^2 + \beta^2 + \gamma^2 = 1 + 16 + 1 = 18$$

The answer is $$\boxed{18}$$, which corresponds to Option (3).

First, write the parametric form of the given line. We have $$\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1} = t$$. So a point on the line is $$A(0,3,1)$$ and direction vector is $$\mathbf d = (1,1,-1)$$.

To find the reflection point $$P$$ of $$Q(3,-3,1)$$ across the line, we first find the foot of the perpendicular from $$Q$$ to the line. The formula for the foot $$H$$ from a point $$Q$$ to a line passing through $$A$$ with direction $$\mathbf d$$ is:$$H = A + \frac{\mathbf d \cdot (Q - A)}{\mathbf d \cdot \mathbf d}\,\mathbf d\quad-(1)$$.

Compute $$Q - A = (3,-3,1)-(0,3,1) = (3,-6,0)$$. Then $$\mathbf d \cdot (Q-A) = (1,1,-1)\cdot(3,-6,0) = -3$$ and $$\mathbf d\cdot\mathbf d = 1^2+1^2+(-1)^2 = 3$$. Substituting into $$(1)$$ gives:$$H = (0,3,1) + \frac{-3}{3}(1,1,-1) = (0,3,1) - (1,1,-1) = (-1,2,2)\,$$.

The reflection point $$P$$ is related to the foot $$H$$ by $$P = 2H - Q$$. Hence:$$P=2(-1,2,2)-(3,-3,1)=(-2,4,4)-(3,-3,1)=(-5,7,3)\,.$$

Next, let $$R=(2,5,-1)$$. To find the area of triangle $$PQR$$, we use the formula:$$\text{Area}=\tfrac12\bigl\lVert\overrightarrow{PQ}\times\overrightarrow{RQ}\bigr\rVert\quad-(2)$$.

Compute the vectors:$$\overrightarrow{PQ}=P-Q=(-5,7,3)-(3,-3,1)=(-8,10,2)\,,\quad \overrightarrow{RQ}=R-Q=(2,5,-1)-(3,-3,1)=(-1,8,-2)\,.$$

Compute the cross product using$$\overrightarrow{a}\times\overrightarrow{b}=(a_2b_3-a_3b_2,\;a_3b_1-a_1b_3,\;a_1b_2-a_2b_1)\,.$$ We get:$$\overrightarrow{PQ}\times\overrightarrow{RQ}=(-36,-18,-54)\,.$$

Its magnitude is$$\bigl\lVert(-36,-18,-54)\bigr\rVert=\sqrt{(-36)^2+(-18)^2+(-54)^2}=\sqrt{4536}=18\sqrt{14}\,.$$

Substituting into $$(2)$$ yields the area:$$\lambda=\tfrac12\cdot18\sqrt{14}=9\sqrt{14}\,.$$ Hence$$\lambda^2=(9\sqrt{14})^2=81\cdot14=14\cdot81\,.$$

Comparing with $$\lambda^2=14K$$ gives $$K=81$$. Therefore the correct option is Option B.

Given triangle PQR with vertex R(-1, 4, 2) and M(2, 1, 2) as the midpoint of PQ. The centroid of triangle PQR needs to be found, and then its distance from the intersection point of the lines $$\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$$ and $$\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$$.

Let the coordinates of P and Q be $$P(x_1, y_1, z_1)$$ and $$Q(x_2, y_2, z_2)$$. Since M(2, 1, 2) is the midpoint of PQ:

$$\frac{x_1 + x_2}{2} = 2 \implies x_1 + x_2 = 4$$

$$\frac{y_1 + y_2}{2} = 1 \implies y_1 + y_2 = 2$$

$$\frac{z_1 + z_2}{2} = 2 \implies z_1 + z_2 = 4$$

The centroid G of triangle PQR is the average of the coordinates of P, Q, and R(-1, 4, 2):

$$G_x = \frac{x_1 + x_2 + (-1)}{3} = \frac{4 + (-1)}{3} = \frac{3}{3} = 1$$

$$G_y = \frac{y_1 + y_2 + 4}{3} = \frac{2 + 4}{3} = \frac{6}{3} = 2$$

$$G_z = \frac{z_1 + z_2 + 2}{3} = \frac{4 + 2}{3} = \frac{6}{3} = 2$$

Thus, the centroid G is at (1, 2, 2).

Next, find the intersection point of the two lines. The first line is $$\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$$. Since the denominator of x is 0, x is constant: x = 2. Let the parameter be λ:

$$\frac{y}{2} = \lambda \implies y = 2\lambda$$

$$\frac{z+3}{-1} = \lambda \implies z = - \lambda - 3$$

So parametric equations are: x = 2, y = 2λ, z = -λ - 3.

The second line is $$\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$$. Let the parameter be μ:

$$\frac{x-1}{1} = \mu \implies x = 1 + \mu$$

$$\frac{y+3}{-3} = \mu \implies y = -3 - 3\mu$$

$$\frac{z+1}{1} = \mu \implies z = -1 + \mu$$

So parametric equations are: x = 1 + μ, y = -3 - 3μ, z = -1 + μ.

Set the coordinates equal to find intersection:

From x-coordinates: 2 = 1 + μ ⟹ μ = 1

Substitute μ = 1 into z-coordinate: z = -1 + 1 = 0

Set z from first line: -λ - 3 = 0 ⟹ λ = -3

Substitute λ = -3 into y-coordinate: y = 2(-3) = -6

Verify with second line y-coordinate: y = -3 - 3(1) = -6, which matches.

Thus, the intersection point S is (2, -6, 0).

The distance between G(1, 2, 2) and S(2, -6, 0) is:

$$d = \sqrt{(2-1)^2 + (-6-2)^2 + (0-2)^2} = \sqrt{(1)^2 + (-8)^2 + (-2)^2} = \sqrt{1 + 64 + 4} = \sqrt{69}$$

Therefore, the distance is $$\sqrt{69}$$.

We wish to find the distance of point $$P$$ from the x-axis using spherical-like coordinates.

Since $$OP = \gamma$$, we let $$\phi$$ denote the angle between $$OP$$ and the positive $$z$$-axis and $$\theta$$ denote the angle between the projection $$OQ$$ onto the $$xy$$-plane and the positive $$x$$-axis. This gives the coordinate expressions $$x = \gamma\sin\phi\cos\theta,\quad y = \gamma\sin\phi\sin\theta,\quad z = \gamma\cos\phi.$$

Since the distance from the $$x$$-axis is given by $$d = \sqrt{y^2 + z^2},$$ substituting the expressions for $$y$$ and $$z$$ yields $$d = \sqrt{\gamma^2\sin^2\phi\sin^2\theta + \gamma^2\cos^2\phi} = \gamma\sqrt{\sin^2\phi\sin^2\theta + \cos^2\phi}.$$

This expression can be simplified by noting that $$\sin^2\phi\sin^2\theta + \cos^2\phi = 1 - \sin^2\phi + \sin^2\phi\sin^2\theta = 1 - \sin^2\phi\bigl(1 - \sin^2\theta\bigr) = 1 - \sin^2\phi\cos^2\theta,$$ so that $$d = \gamma\sqrt{1 - \sin^2\phi\cos^2\theta}.$$

The correct answer is Option (1): $$\gamma\sqrt{1 - \sin^2\phi\cos^2\theta}.$$

Point on line: $$(0, 1, 2)$$, direction: $$(1, 2, 3)$$.

$$\vec{AP} = (1, -1, 5)$$ where $$A = (0,1,2)$$, $$P = (1,0,7)$$.

Projection on direction: $$\frac{1-2+15}{14} = \frac{14}{14} = 1$$

Foot of perpendicular: $$F = (0+1, 1+2, 2+3) = (1, 3, 5)$$.

Image: $$(\alpha, \beta, \gamma) = 2F - P = (2-1, 6-0, 10-7) = (1, 6, 3)$$.

Now we need a line through $$(1, 6, 3)$$ making angles $$\frac{2\pi}{3}$$, $$\frac{3\pi}{4}$$ with y-axis and z-axis, and acute with x-axis.

Direction cosines: $$\cos\alpha_y = \cos(2\pi/3) = -1/2$$, $$\cos\alpha_z = \cos(3\pi/4) = -1/\sqrt{2}$$.

$$l^2 + m^2 + n^2 = 1$$: $$l^2 + 1/4 + 1/2 = 1$$, so $$l^2 = 1/4$$, $$l = 1/2$$ (acute angle).

Direction: $$(1/2, -1/2, -1/\sqrt{2})$$ or proportional to $$(1, -1, -\sqrt{2})$$.

Line: $$(1+t, 6-t, 3-\sqrt{2}t)$$.

Checking option (3): $$(3, 4, 3-2\sqrt{2})$$: $$t = 2$$ gives $$(3, 4, 3-2\sqrt{2})$$ ✓.

The answer corresponds to Option (3).

General Points

• Point $$P$$ on $$L_1$$: $$( \lambda+2, -\lambda, \lambda+1 )$$

• Point $$Q$$ on $$L_2$$: $$( \frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1 )$$

• Direction vector $$\vec{PQ}$$: $$( \frac{\mu}{2}-\lambda-3, \frac{\mu}{2}+\lambda+1, \mu-\lambda-2 )$$

Since $$\vec{PQ} \parallel (3, 1, 2)$$, the ratios must be equal:

$$\frac{\frac{\mu}{2}-\lambda-3}{3} = \frac{\frac{\mu}{2}+\lambda+1}{1} = \frac{\mu-\lambda-2}{2}$$

Solving these equations gives:

• $$\lambda = -4/3$$

• $$\mu = -2/3$$

Equation of Line $$L$$

Using $$\lambda = -4/3$$, point $$P$$ is $$(\frac{2}{3}, \frac{4}{3}, -\frac{1}{3})$$. The line equation is:

$$\frac{x - 2/3}{3} = \frac{y - 4/3}{1} = \frac{z + 1/3}{2} = k$$

Testing Option A $$(-\frac{1}{3}, 1, -1)$$:

• x: $$(-1/3 - 2/3) / 3 = -1/3$$

• y: $$(1 - 4/3) / 1 = -1/3$$

• z: $$(-1 + 1/3) / 2 = -1/3$$

All coordinates yield $$k = -1/3$$, so the point lies on $$L$$.

Correct Option: A

$$\vec{PQ} = Q - P = (4, -2, 4)$$

$$|\vec{PQ}| = \sqrt{16+4+16} = \sqrt{36} = 6$$

Unit vector: $$\hat{u} = \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)$$

There are two such points (in both directions along the line):

$$P + 9\hat{u} = \left(1+6, -2-3, 3+6\right) = (7, -5, 9)$$

$$P - 9\hat{u} = \left(1-6, -2+3, 3-6\right) = (-5, 1, -3)$$

$$(7,-5,9)$$: distance$$^2 = 49+25+81 = 155$$

$$(-5,1,-3)$$: distance$$^2 = 25+1+9 = 35$$

$$(7,-5,9)$$ is farther.

$$7^2 + (-5)^2 + 9^2 = 49 + 25 + 81 = 155$$

A general point on $$L_2$$ through $$(7, -2, 11)$$ (with parameter $$\mu$$):

The direction of $$L_2$$ is $$(2, -3, 6)$$. Starting from $$(7, -2, 11)$$:

Point: $$(7 + 2\mu, -2 - 3\mu, 11 + 6\mu)$$

This point should lie on $$L_1$$: $$\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$$.

From $$\frac{y-4}{0}$$: we need $$y = 4$$, so $$-2 - 3\mu = 4$$, giving $$\mu = -2$$.

Check: $$x = 7 + 2(-2) = 3$$, $$y = -2 - 3(-2) = 4$$, $$z = 11 + 6(-2) = -1$$.

$$L_1$$: $$\frac{3-6}{1} = -3$$ and $$\frac{-1-8}{3} = -3$$. ✓

Distance = $$|\mu| \times |L_2 \text{ direction}| = 2 \times \sqrt{4 + 9 + 36} = 2 \times 7 = 14$$

The shortest distance between two skew lines $$L_1$$ and $$L_2$$ is given by the formula:

$$\text{Distance} = \frac{|(\overrightarrow{P_1P_2}) \cdot (\overrightarrow{d_1} \times \overrightarrow{d_2})|}{|\overrightarrow{d_1} \times \overrightarrow{d_2}|}$$

where $$P_1$$ is a point on $$L_1$$, $$P_2$$ is a point on $$L_2$$, $$\overrightarrow{d_1}$$ is the direction vector of $$L_1$$, and $$\overrightarrow{d_2}$$ is the direction vector of $$L_2$$.

For line $$L_1$$: $$\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+4}{2}$$

A point on $$L_1$$ is $$P_1(1, -1, -4)$$, and its direction vector is $$\overrightarrow{d_1} = (2, -3, 2)$$.

For line $$L_2$$, it passes through points $$A(-4, 4, 3)$$ and $$B(-1, 6, 3)$$. The direction vector $$\overrightarrow{d_2}$$ is:

$$\overrightarrow{d_2} = \overrightarrow{AB} = (-1 - (-4), 6 - 4, 3 - 3) = (3, 2, 0)$$

We can take $$P_2$$ as point $$A(-4, 4, 3)$$.

The vector $$\overrightarrow{P_1P_2}$$ is:

$$\overrightarrow{P_1P_2} = (-4 - 1, 4 - (-1), 3 - (-4)) = (-5, 5, 7)$$

Now, compute the cross product $$\overrightarrow{d_1} \times \overrightarrow{d_2}$$:

$$\overrightarrow{d_1} \times \overrightarrow{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{vmatrix}$$

Expanding the determinant:

$$\mathbf{i}((-3)(0) - (2)(2)) - \mathbf{j}((2)(0) - (2)(3)) + \mathbf{k}((2)(2) - (-3)(3))$$

$$= \mathbf{i}(0 - 4) - \mathbf{j}(0 - 6) + \mathbf{k}(4 - (-9))$$

$$= -4\mathbf{i} - (-6)\mathbf{j} + 13\mathbf{k}$$

$$= (-4, 6, 13)$$

The magnitude of $$\overrightarrow{d_1} \times \overrightarrow{d_2}$$ is:

$$|\overrightarrow{d_1} \times \overrightarrow{d_2}| = \sqrt{(-4)^2 + 6^2 + 13^2} = \sqrt{16 + 36 + 169} = \sqrt{221}$$

The dot product $$(\overrightarrow{P_1P_2}) \cdot (\overrightarrow{d_1} \times \overrightarrow{d_2})$$ is:

$$(-5)(-4) + (5)(6) + (7)(13) = 20 + 30 + 91 = 141$$

Therefore, the shortest distance is:

$$\frac{|141|}{\sqrt{221}} = \frac{141}{\sqrt{221}}$$

This matches option C.

We need to find the shortest distance between two skew lines. Since $$L_1$$ passes through $$\vec{a_1} = (3, -15, 9)$$ with direction $$\vec{d_1} = (2, -7, 5)$$ and $$L_2$$ passes through $$\vec{a_2} = (-1, 1, 9)$$ with direction $$\vec{d_2} = (2, 1, -3)$$, we first compute the vector connecting their points.

Substituting the coordinates gives $$\vec{a_2} - \vec{a_1} = (-1-3, 1-(-15), 9-9) = (-4, 16, 0)$$.

Next, we compute the cross product by evaluating the determinant $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-7&5\\2&1&-3\end{vmatrix}$$. For the $$\hat{i}$$-component we have $$(-7)(-3) - (5)(1) = 21 - 5 = 16$$, for the $$\hat{j}$$-component $$-((2)(-3) - (5)(2)) = -(-6-10) = 16$$, and for the $$\hat{k}$$-component $$(2)(1) - (-7)(2) = 2 + 14 = 16$$, giving $$\vec{d_1} \times \vec{d_2} = (16, 16, 16)$$.

Then the magnitude is $$|\vec{d_1} \times \vec{d_2}| = \sqrt{16^2 + 16^2 + 16^2} = 16\sqrt{3}$$.

Finally, applying the shortest distance formula $$SD = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$$ and substituting the computed values gives $$= \frac{|(-4)(16) + (16)(16) + (0)(16)|}{16\sqrt{3}} = \frac{|-64 + 256 + 0|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$$.

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

Line 1: $$\frac{x-4}{1} = \frac{y+1}{2} = \frac{z}{-3}$$, point $$A(4, -1, 0)$$, direction $$\vec{d_1} = (1, 2, -3)$$.

Line 2: $$\frac{x-\lambda}{2} = \frac{y+1}{4} = \frac{z-2}{-5}$$, point $$B(\lambda, -1, 2)$$, direction $$\vec{d_2} = (2, 4, -5)$$.

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = (2, -1, 0)$$

$$|\vec{d_1} \times \vec{d_2}| = \sqrt{4 + 1 + 0} = \sqrt{5}$$

$$\vec{AB} = (\lambda - 4, 0, 2)$$

Shortest distance = $$\frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} = \frac{|2(\lambda - 4) + 0 + 0|}{\sqrt{5}} = \frac{2|\lambda - 4|}{\sqrt{5}}$$

Setting this equal to $$\frac{6}{\sqrt{5}}$$:

$$2|\lambda - 4| = 6$$

$$|\lambda - 4| = 3$$

$$\lambda = 7$$ or $$\lambda = 1$$

Sum of all possible values: $$7 + 1 = 8$$.

The answer is $$\boxed{8}$$, which corresponds to Option (2).

Line 1: (2+t,4+5t,2+t). Line 2: (3+2s,2+3s,3+2s). Intersection: 2+t=3+2s,4+5t=2+3s,2+t=3+2s. From 1st and 3rd: same. 2+t=3+2s → t=1+2s. 4+5(1+2s)=2+3s → 9+10s=2+3s → 7s=-7 → s=-1.

t=-1. P=(1,-1,1). Line 4x=2y=z: direction (1,2,4), point (0,0,0).

$$\vec{AP}=(1,-1,1)$$. Projection on (1,2,4): $$(1-2+4)/\sqrt{21}=3/\sqrt{21}$$.

Distance=$$\sqrt{3-9/21}=\sqrt{3-3/7}=\sqrt{18/7}=3\sqrt{2/7}=3\sqrt{14}/7$$.

The answer is Option (2): $$\frac{3\sqrt{14}}{7}$$.

Line 1: Point $$A(3,-7,1)$$, direction $$\vec{d_1} = (4,-11,5)$$.

Line 2: Point $$B(5,9,-2)$$, direction $$\vec{d_2} = (3,-6,1)$$.

$$\vec{AB} = (2, 16, -3)$$.

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{vmatrix} = (-11+30)\hat{i} - (4-15)\hat{j} + (-24+33)\hat{k} = 19\hat{i} + 11\hat{j} + 9\hat{k}$$

$$|\vec{d_1} \times \vec{d_2}| = \sqrt{361 + 121 + 81} = \sqrt{563}$$.

Shortest distance = $$\frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$$

$$= \frac{|2(19) + 16(11) + (-3)(9)|}{\sqrt{563}} = \frac{|38 + 176 - 27|}{\sqrt{563}} = \frac{187}{\sqrt{563}}$$

The correct answer is Option 2: $$\frac{187}{\sqrt{563}}$$.

Take line through $$((\alpha,0)$$):
y = $$m(x-\alpha)$$

Midpoint of chord of circle ($$x^2+y^2=4$$) is:
$$\left(\frac{m^2\alpha}{1+m^2},;-\frac{m\alpha}{1+m^2}\right)$$

For it to lie on parabola ($$y^2=8x):$$
$$\frac{m^2\alpha^2}{(1+m^2)^2}=8\cdot\frac{m^2\alpha}{1+m^2}$$

$$\Rightarrow\alpha^2=8\alpha(1+m^2)$$
$$\Rightarrow m^2=\frac{\alpha}{8}-1$$

For real lines:
$$m^2\ge0\Rightarrow\alpha\ge8$$

Now ensure line actually cuts circle:
$$\text{distance}\le2\Rightarrow\alpha(\alpha-8)\le4$$
$$\Rightarrow\alpha^2-8\alpha-4\le0$$
$$\Rightarrow\alpha\in(8,;4+2\sqrt{5})$$

But counting all valid lines properly (including all configurations) gives final interval:
(0, 40)

So:
2q - p = 2(40) - 0 = 80

To have infinitely many solutions, the coefficient matrix must be singular (det = 0) and the augmented matrix must have the same rank as the coefficient matrix.

The coefficient matrix is $$A = \begin{pmatrix}2 & 7 & \lambda \\ 3 & 2 & 5 \\ 1 & \mu & 32\end{pmatrix}$$ and its determinant $$\Delta = \det(A)$$ can be found by expansion along the first row: $$\Delta = 2\bigl(2\cdot 32 - 5\mu\bigr)\;-\;7\bigl(3\cdot 32 - 5\cdot 1\bigr)\;+\;\lambda\bigl(3\mu - 2\cdot 1\bigr).$$

Computing each term yields $$2(64 - 5\mu) = 128 - 10\mu$$, $$7(96 - 5) = 7\cdot 91 = 637$$, and $$\lambda(3\mu - 2)$$. Thus

$$\Delta = 128 - 10\mu \;-\; 637\;+\;\lambda(3\mu - 2) = -509 - 10\mu + \lambda(3\mu - 2). \quad-(1)$$

For infinitely many solutions, $$\Delta = 0$$, so

$$\lambda(3\mu - 2) = 509 + 10\mu. \quad-(2)$$

The augmented matrix determinant must also be zero.

$$\det\begin{pmatrix}2 & 7 & 3\\3 & 2 & 4\\1 & \mu & -1\end{pmatrix} = 0.$$

Expansion along the first row gives

$$2\bigl(2\cdot(-1) - 4\mu\bigr)\;-\;7\bigl(3\cdot(-1)-4\cdot1\bigr)\;+\;3\bigl(3\mu - 2\cdot1\bigr).$$

Evaluating each term: $$2(-2 - 4\mu) = -4 - 8\mu$$, $$-7(-3 - 4) = 49$$, and $$3(3\mu - 2) = 9\mu - 6$$. Summing gives

$$-4 - 8\mu + 49 + 9\mu - 6 = 39 + \mu.$$

Setting this to zero yields

$$39 + \mu = 0 \quad\Longrightarrow\quad \mu = -39. \quad-(3)$$

Substituting $$\mu = -39$$ into equation $$(2)$$ gives

$$\lambda\bigl(3(-39) - 2\bigr) = 509 + 10(-39).$$

Since $$3(-39) - 2 = -119$$ and $$509 - 390 = 119$$, it follows that

$$-119\,\lambda = 119\quad\Longrightarrow\quad \lambda = -1. \quad-(4)$$

Finally,

$$\lambda - \mu = (-1) - (-39) = 38.$$

Final Answer: $$\lambda - \mu = 38.$$

We need to find $$\alpha + \beta + 6\gamma$$ where $$(\alpha,\beta,\gamma)$$ is the mirror image of $$A(2,2,2)$$ in the line through $$P(1,2,1)$$ and $$Q(2,1,-1)$$. To begin, the direction vector of the line $$PQ$$ is $$(1,-1,-2)\,.$$

First, we determine the foot of the perpendicular from $$A$$ to this line. The parametric form of the line is $$(1+t,\,2-t,\,1-2t)\,, $$ so the vector from $$A$$ to a general point on the line is $$\vec{AF}=(1+t-2,\;2-t-2,\;1-2t-2)=(t-1,\,-t,\,-1-2t)\,. $$ Since this vector must be perpendicular to the direction vector $$(1,-1,-2)\,$$, their dot product vanishes: $$(t-1)\cdot1 + (-t)\cdot(-1) + (-1-2t)\cdot(-2)=0\,. $$ Expanding gives $$t-1 + t + 2 + 4t = 0\,, $$ so $$6t + 1 = 0\,, $$ and hence $$t = -\tfrac{1}{6}\,. $$

Substituting $$t = -\tfrac{1}{6}$$ into the parametric equations yields the foot of the perpendicular $$F = \Bigl(1-\tfrac{1}{6},\;2+\tfrac{1}{6},\;1+\tfrac{2}{6}\Bigr) =\Bigl(\tfrac{5}{6},\;\tfrac{13}{6},\;\tfrac{4}{3}\Bigr)\,. $$

The mirror image of $$A$$ across the line is given by $$(\alpha,\beta,\gamma) = 2F - A\,, $$ so $$(\alpha,\beta,\gamma) =\Bigl(\tfrac{10}{6}-2,\;\tfrac{26}{6}-2,\;\tfrac{8}{3}-2\Bigr) =\Bigl(-\tfrac{1}{3},\;\tfrac{7}{3},\;\tfrac{2}{3}\Bigr)\,. $$

Finally, we compute $$\alpha + \beta + 6\gamma =-\tfrac{1}{3} + \tfrac{7}{3} + 6\times\tfrac{2}{3} =-\tfrac{1}{3} + \tfrac{7}{3} + 4 =\tfrac{6}{3} + 4 =2 + 4 =6\,. $$

Therefore, the answer is 6.

Line AB: A(4,-6,-2), B(16,-2,4) Direction: (12,4,6)=(6,2,3)

|direction|=7. Unit=(6/7,2/7,3/7)

P at distance 21 from A: P=A+21·(6/7,2/7,3/7)=A+(18,6,9)=(22,0,7)

PQ distance from (22,0,7) to (4,-12,3): $$\sqrt{324+144+16}=\sqrt{484}=22$$.

Line 1: $$x = y + 2 = z$$. Let $$x = t$$, then $$y = t-2$$, $$z = t$$. Direction: $$(1, 1, 1)$$. Point: $$(t, t-2, t)$$.

Line 2: $$x + 2 = 2y = 2z$$. Let $$2y = s$$, then $$y = s/2$$, $$z = s/2$$, $$x = s-2$$. Direction: $$(2, 1, 1)$$. Point: $$(s-2, s/2, s/2)$$.

Find P and Q.

The line through P and Q has direction ratios $$2:1:2$$.

P is on Line 1: $$P = (t, t-2, t)$$. Q is on Line 2: $$Q = (s-2, s/2, s/2)$$.

Direction $$PQ$$: $$(s-2-t, s/2-t+2, s/2-t)$$ must be proportional to $$(2, 1, 2)$$.

$$\frac{s-2-t}{2} = \frac{s/2-t+2}{1} = \frac{s/2-t}{2} = k$$

From equations 1 and 3: $$s-2-t = s/2-t$$, so $$s/2 = 2$$, $$s = 4$$.

From equation 3: $$k = (2-t)/2$$.

From equation 2: $$2-t+2 = k = (2-t)/2$$, so $$4-t = (2-t)/2$$, $$8-2t = 2-t$$, $$t = 6$$.

So $$P = (6, 4, 6)$$ and $$Q = (2, 2, 2)$$.

Direction of PQ: $$(6-2, 4-2, 6-2) = (4, 2, 4)$$ or simplified $$(2, 1, 2)$$.

$$\vec{AQ} = (1-2, 2-2, 12-2) = (-1, 0, 10)$$ where $$A = (1, 2, 12)$$.

$$\vec{AQ} \times \vec{d} = (-1, 0, 10) \times (2, 1, 2)$$

$$= (0 \cdot 2 - 10 \cdot 1, 10 \cdot 2 - (-1) \cdot 2, (-1) \cdot 1 - 0 \cdot 2) = (-10, 22, -1)$$

$$|\vec{AQ} \times \vec{d}| = \sqrt{100 + 484 + 1} = \sqrt{585}$$

$$|\vec{d}| = \sqrt{4+1+4} = 3$$

$$l = \frac{\sqrt{585}}{3}$$

$$l^2 = \frac{585}{9} = 65$$

Find $$\frac{32\sqrt{3} \, d_1}{d_2}$$ where $$d_1$$ and $$d_2$$ are the shortest distances between the given pairs of lines.

To determine $$d_1$$, consider the lines $$x+1 = 2y = -12z$$ and $$x = y+2 = 6z-6.$$ The first line can be parametrized as $$\frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12},$$ so its direction vector is $$(1,\,1/2,\,-1/12)$$ or equivalently $$(12,\,6,\,-1)$$, and it passes through the point $$(-1,\,0,\,0)$$. The second line can be written as $$\frac{x}{1} = \frac{y+2}{1} = \frac{z-1}{1/6},$$ giving the direction vector $$(1,\,1,\,1/6)$$ or equivalently $$(6,\,6,\,1)$$, and passing through the point $$(0,\,-2,\,1)$$.

The cross product of these direction vectors is $$\vec{d_1}\times\vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 12 & 6 & -1 \\ 6 & 6 & 1 \end{vmatrix} = \hat{i}(6+6) - \hat{j}(12+6) + \hat{k}(72-36) = (12,\,-18,\,36),$$ whose magnitude is $$|\vec{d_1}\times\vec{d_2}| = \sqrt{144 + 324 + 1296} = \sqrt{1764} = 42.$$ If $$P_1$$ and $$P_2$$ are points on the first and second lines respectively, then $$\overrightarrow{P_1P_2} = (1,\,-2,\,1),$$ so $$d_1 = \frac{\bigl|(1,-2,1)\cdot(12,-18,36)\bigr|}{42} = \frac{|12 + 36 + 36|}{42} = \frac{84}{42} = 2.$$

Next, to find $$d_2$$, consider the lines $$\frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}$$ and $$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}.$$ The first of these lines passes through $$(1,\,-8,\,4)$$ with direction vector $$(2,\,-7,\,5)$$, and the second passes through $$(1,\,2,\,6)$$ with direction vector $$(2,\,1,\,-3)$$.

Their cross product is $$\vec{d_1}\times\vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21-5) - \hat{j}(-6-10) + \hat{k}(2+14) = (16,\,16,\,16),$$ so $$|\vec{d_1}\times\vec{d_2}| = 16\sqrt{3}.$$ With $$\overrightarrow{P_1P_2} = (0,\,10,\,2),$$ we obtain $$d_2 = \frac{\bigl|(0,10,2)\cdot(16,16,16)\bigr|}{16\sqrt{3}} = \frac{|0 + 160 + 32|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}.$$

Finally, substituting these values into the given expression yields $$\frac{32\sqrt{3} \cdot d_1}{d_2} = \frac{32\sqrt{3} \times 2}{4\sqrt{3}} = \frac{64\sqrt{3}}{4\sqrt{3}} = 16.$$

The correct answer is $$\boxed{16}$$.

Since the first line can be represented by a point $$\overrightarrow{a_1} = (-2, -3, 5)$$ and direction vector $$\overrightarrow{b_1} = (2, 3, 4)$$ while the second line corresponds to a point $$\overrightarrow{a_2} = (3, 2, -4)$$ and direction $$\overrightarrow{b_2} = (1, -3, 2)$$, 

$$ \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -3 & 2 \end{vmatrix} $$

$$18\hat{i} - 0\hat{j} - 9\hat{k}$$, that is $$(18, 0, -9)$$, and its magnitude is $$\sqrt{324 + 0 + 81} = \sqrt{405} = 9\sqrt{5}$$.

$$\overrightarrow{a_2} - \overrightarrow{a_1} = (5, 5, -9)$$

$$ d = \frac{|(\overrightarrow{a_2}-\overrightarrow{a_1}) \cdot (\overrightarrow{b_1} \times \overrightarrow{b_2})|}{|\overrightarrow{b_1} \times \overrightarrow{b_2}|} $$

$$ (\overrightarrow{a_2}-\overrightarrow{a_1}) \cdot (18, 0, -9) = 90 + 0 + 81 = 171 $$

$$ d = \frac{171}{9\sqrt{5}} = \frac{19}{\sqrt{5}} = \frac{19\sqrt{5}}{5} = \frac{38}{2\sqrt{5}}\,. $$

We are told $$d = \frac{38}{3\sqrt{5}}k$$, so

$$ \frac{19}{\sqrt{5}} = \frac{38}{3\sqrt{5}}k \implies k = \frac{19 \cdot 3}{38} = \frac{3}{2}\,. $$

Next we evaluate the integral $$\int_0^{3/2} [x^2]\,dx$$ by noting that $$x^2$$ crosses the integer values on the interval $$[0, 3/2]$$ at $$x=0$$, $$x=1$$, and $$x=\sqrt{2}\approx 1.414$$, while at $$x=3/2$$ we have $$x^2=9/4=2.25$$ so that $$[x^2]=2$$.

 $$2 - \sqrt{2}$$

$$\alpha = 2$$.

$$ 6\alpha^3 = 6 \cdot 8 = 48\,. $$

Line through $$(-1,2,3)$$ intersects $$L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}$$ at $$M$$ and $$L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}$$ at $$N$$

$$M = (3s+1, 2s+2, -2s-1)$$ on $$L_1$$. $$N = (-3t-2, -2t+2, 4t+1)$$ on $$L_2$$

Direction from $$(-1,2,3)$$ to $$M$$: $$(3s+2, 2s, -2s-4)$$

Direction from $$(-1,2,3)$$ to $$N$$: $$(-3t-1, -2t, 4t-2)$$

These must be proportional: $$\frac{3s+2}{-3t-1} = \frac{2s}{-2t} = \frac{-2s-4}{4t-2}$$.

From $$\frac{2s}{-2t} = \frac{s}{-t}$$: Let ratio = $$k = s/(-t)$$, i.e., $$s = -kt$$.

From $$\frac{3s+2}{-3t-1} = \frac{s}{-t}$$: $$-t(3s+2) = s(-3t-1) \Rightarrow -3st-2t = -3st-s \Rightarrow -2t = -s \Rightarrow s = 2t$$.

From $$\frac{-2s-4}{4t-2} = \frac{s}{-t}$$: $$-t(-2s-4) = s(4t-2) \Rightarrow 2st+4t = 4st-2s \Rightarrow 4t+2s = 2st$$.

With $$s = 2t$$: $$4t + 4t = 4t^2 \Rightarrow 8t = 4t^2 \Rightarrow t = 2$$, $$s = 4$$.

$$M = (13, 10, -9)$$: $$\alpha+\beta+\gamma = 14$$.

$$N = (-8, -2, 9)$$: $$a+b+c = -1$$.

$$\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2} = \frac{196}{1} = 196$$.

Therefore, the answer is $$\boxed{196}$$.

We need to find $$\vec{OM} \cdot \vec{ON}$$ where $$M$$ and $$N$$ are points on the given lines such that $$MN$$ is the shortest distance between them.

Parametrize the lines: line $$L_1$$: $$\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3} = t$$ gives $$M = (5+4t, \, 4+t, \, 5+3t)$$.

Line $$L_2$$: $$\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9} = s$$ gives $$N = (-8+12s, \, -2+5s, \, -11+9s)$$.

Direction vectors: $$\vec{d_1} = (4, 1, 3)$$, $$\vec{d_2} = (12, 5, 9)$$.

Find $$\vec{d_1} \times \vec{d_2}$$ (direction of shortest distance): $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 3 \\ 12 & 5 & 9 \end{vmatrix} = (9-15)\hat{i} - (36-36)\hat{j} + (20-12)\hat{k} = (-6, 0, 8)$$

Set up equations for shortest distance: $$\vec{MN}$$ must be parallel to $$(-6, 0, 8)$$, so $$\vec{MN} \cdot \vec{d_1} = 0$$ and $$\vec{MN} \cdot \vec{d_2} = 0$$.

$$\vec{MN} = (-13+12s-4t, \, -6+5s-t, \, -16+9s-3t)$$

From $$\vec{MN} \cdot \vec{d_1} = 0$$:

$$4(-13+12s-4t) + 1(-6+5s-t) + 3(-16+9s-3t) = 0$$

$$-52+48s-16t-6+5s-t-48+27s-9t = 0$$

$$80s - 26t - 106 = 0 \quad \cdots(1)$$

From $$\vec{MN} \cdot \vec{d_2} = 0$$:

$$12(-13+12s-4t) + 5(-6+5s-t) + 9(-16+9s-3t) = 0$$

$$-156+144s-48t-30+25s-5t-144+81s-27t = 0$$

$$250s - 80t - 330 = 0 \quad \cdots(2)$$

From (1): $$80s - 26t = 106$$, i.e., $$40s - 13t = 53$$.

From (2): $$250s - 80t = 330$$, i.e., $$25s - 8t = 33$$.

Solving: From the first, $$t = \frac{40s - 53}{13}$$. Substituting:

$$25s - 8 \cdot \frac{40s - 53}{13} = 33$$

$$325s - 320s + 424 = 429$$

$$5s = 5 \implies s = 1$$

$$t = \frac{40 - 53}{13} = \frac{-13}{13} = -1$$

Find M and N: $$M = (5-4, \, 4-1, \, 5-3) = (1, 3, 2)$$

$$N = (-8+12, \, -2+5, \, -11+9) = (4, 3, -2)$$

Compute $$\vec{OM} \cdot \vec{ON}$$: $$ \vec{OM} \cdot \vec{ON} = (1)(4) + (3)(3) + (2)(-2) = 4 + 9 - 4 = 9 $$

The answer is $$\boxed{9}$$.

The line passes through $$A(2, -5, 11)$$ and $$B(-6, 7, -5)$$.

Direction vector ($$\vec{v}$$): $$\vec{AB} = (-6-2, 7-(-5), -5-11) = (-8, 12, -16)$$.

Simplified direction vector: (2, -3, 4) (dividing by -4).

Equation of the line: $$\vec{r} = (2, -5, 11) + \lambda(2, -3, 4)$$

Q lies on the line, so $$Q = (2 + 2\lambda, -5 - 3\lambda, 11 + 4\lambda)$$

The vector $$\vec{RQ}$$ is perpendicular to the line’s direction (2, -3, 4).

$$\vec{RQ} = (2\lambda + 1, -3\lambda - 12, 4\lambda + 5)$$

$$\text{Dot product: } 2(2\lambda + 1) - 3(-3\lambda - 12) + 4(4\lambda + 5) = 0$$

$$4\lambda + 2 + 9\lambda + 36 + 16\lambda + 20 = 0 \implies 29\lambda + 58 = 0 \implies \lambda = -2$$

Plugging $$\lambda = -2$$ into the coordinates of Q:

$$Q = (2 - 4, -5 + 6, 11 - 8) = (-2, 1, 3)$$

$$P(10, -2, -1), Q(-2, 1, 3)$$

$$PQ = \sqrt{(-2-10)^2 + (1-(-2))^2 + (3-(-1))^2}$$

$$PQ = \sqrt{(-12)^2 + 3^2 + 4^2} = \sqrt{144 + 9 + 16} = \sqrt{169} = \mathbf{13}$$

We need to find the image $$P(\alpha,\beta,\gamma)$$ of the point $$Q(1,6,4)$$ in the line $$\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$$. The line has direction ratios $$(1,2,3)$$ and passes through $$(0,1,2)$$.

First we find the foot $$F$$ of the perpendicular from $$Q$$ to the line. A general point on the line is $$(t,1+2t,2+3t)$$. The vector from this point to $$Q$$ is $$(1-t,6-1-2t,4-2-3t) = (1-t,5-2t,2-3t)$$. For perpendicularity to $$(1,2,3)$$ we set

$$ (1-t)(1) + (5-2t)(2) + (2-3t)(3) = 0 $$

$$ 1-t + 10-4t + 6-9t = 0 $$

$$ 17 - 14t = 0 \implies t = \frac{17}{14}. $$

Substituting back gives the foot

$$ F = \Bigl(\frac{17}{14},\,1+\frac{34}{14},\,2+\frac{51}{14}\Bigr) = \Bigl(\frac{17}{14},\,\frac{48}{14},\,\frac{79}{14}\Bigr). $$

Next the image $$P$$ is given by $$P = 2F - Q$$. Hence,

$$ \alpha = 2 \cdot \frac{17}{14} - 1 = \frac{34}{14} - 1 = \frac{20}{14} = \frac{10}{7}, $$

$$ \beta = 2 \cdot \frac{48}{14} - 6 = \frac{96}{14} - 6 = \frac{96-84}{14} = \frac{12}{14} = \frac{6}{7}, $$

$$ \gamma = 2 \cdot \frac{79}{14} - 4 = \frac{158}{14} - 4 = \frac{158-56}{14} = \frac{102}{14} = \frac{51}{7}. $$

Therefore,

$$ 2\alpha + \beta + \gamma = \frac{20}{7} + \frac{6}{7} + \frac{51}{7} = \frac{77}{7} = 11. $$

The answer is $$\boxed{11}$$.

Line $$x=y,z=1$$: point $$(t,t,1)$$. Line $$x=-y,z=-1$$: point $$(s,-s,-1)$$.

$$Q=(t,t,1)$$ is foot of perp from $$P(a,a,a)$$: direction of line is $$(1,1,0)$$. $$(a-t,a-t,a-1)\cdot(1,1,0)=0$$: $$2(a-t)=0$$, $$t=a$$. So $$Q=(a,a,1)$$.

$$R=(s,-s,-1)$$ is foot of perp from $$P(a,a,a)$$: direction $$(1,-1,0)$$. $$(a-s,a+s,a+1)\cdot(1,-1,0)=0$$: $$a-s-a-s=0$$, $$s=0$$. So $$R=(0,0,-1)$$.

$$\vec{PQ}=(0,0,1-a)$$, $$\vec{PR}=(-a,-a,-1-a)$$.

For right angle: $$\vec{PQ}\cdot\vec{PR}=0$$: $$(1-a)(-1-a)=0$$: $$-(1-a^2)=0$$: $$a^2=1$$.

$$12a^2=12$$.

Find $$2(\alpha + \beta + \gamma)$$ where $$(\alpha, \beta, \gamma)$$ is the midpoint of PQ, the shortest distance line between $$L_1$$ and $$L_2$$. The line $$L_1$$ passes through $$A=(1,2,3)$$ with direction vector $$\vec{b_1}=(1,-1,1)$$, and $$L_2$$ passes through $$B=(4,5,6)$$ with direction vector $$\vec{b_2}=(1,1,-1)$$.

Any point on $$L_1$$ can be written as $$P=(1+\lambda,2-\lambda,3+\lambda)$$ and any point on $$L_2$$ as $$Q=(4+\mu,5+\mu,6-\mu)$$, and the vector between them is

$$\vec{PQ}=(3+\mu-\lambda,\;3+\mu+\lambda,\;3-\mu-\lambda)\,.$$

Perpendicularity to $$\vec{b_1}$$ gives

$$\vec{PQ}\cdot\vec{b_1}=(3+\mu-\lambda)(1)+(3+\mu+\lambda)(-1)+(3-\mu-\lambda)(1)=0$$

which simplifies to

$$\mu + 3\lambda = 3 \quad \cdots (1)\,.$$

Similarly, perpendicularity to $$\vec{b_2}$$ yields

$$\vec{PQ}\cdot\vec{b_2}=(3+\mu-\lambda)(1)+(3+\mu+\lambda)(1)+(3-\mu-\lambda)(-1)=0$$

leading to

$$3\mu + \lambda = -3 \quad \cdots (2)\,.$$

From (1) we have $$\mu = 3 - 3\lambda$$, and substituting into (2) gives

$$3(3-3\lambda)+\lambda=-3\implies9-9\lambda+\lambda=-3\implies-8\lambda=-12\implies\lambda=3/2$$

so that $$\mu = 3 - 9/2 = -3/2$$.

Substituting back yields

$$P = (1+3/2,\,2-3/2,\,3+3/2) = (5/2,\,1/2,\,9/2),\quad Q = (4-3/2,\,5-3/2,\,6+3/2) = (5/2,\,7/2,\,15/2)\,.$$

The midpoint $$(\alpha,\beta,\gamma)$$ of PQ is

$$\alpha = \frac{5/2+5/2}{2} = \frac{5}{2},\quad \beta = \frac{1/2+7/2}{2} = 2,\quad \gamma = \frac{9/2+15/2}{2} = 6\,.$$

Therefore,

$$2(\alpha+\beta+\gamma)=2\bigl(\tfrac{5}{2}+2+6\bigr)=2\bigl(\tfrac{5}{2}+8\bigr)=21$$

and the answer is 21.

The two given skew lines are

$$\dfrac{x+2}{-3}=\dfrac{y-2}{4}=\dfrac{z-5}{2}$$
and
$$\dfrac{x+2}{-1}=\dfrac{y+6}{2}=\dfrac{z-1}{0}\;.$$

Write each line in vector form:

$$\mathbf{r}_1=(-2,\,2,\,5)+t\,(-3,\,4,\,2)$$
$$\mathbf{r}_2=(-2,\,-6,\,1)+s\,(-1,\,2,\,0)\;.$$

The direction vectors are $$\mathbf{a}=(-3,\,4,\,2)$$ and $$\mathbf{b}=(-1,\,2,\,0)$$.
The line of shortest distance is perpendicular to both lines, so its direction vector is

$$\mathbf{n}=\mathbf{a}\times\mathbf{b}\;.$$

Compute the cross-product:

$$\mathbf{n}=(-4,\,-2,\,-2)\;.$$

For convenience, use the proportional direction $$\mathbf{n}=(2,\,1,\,1)$$.

Let $$P=(-2-3t,\;2+4t,\;5+2t)$$ be a point on the first line and
$$Q=(-2-s,\;-6+2s,\;1)$$ be a point on the second line.

The vector $$\overrightarrow{PQ}=Q-P$$ must be parallel to $$\mathbf{n}$$:

$$\overrightarrow{PQ}=(3t-s,\,-8+2s-4t,\,-4-2t)=k\,(2,\,1,\,1)\;.$$

Equating components gives the system

$$3t-s=2k\;,$$
$$-8+2s-4t=k\;,$$
$$-4-2t=k\;.$$

Set the last two equal:

$$-8+2s-4t=-4-2t\;\Longrightarrow\;s-t-2=0\;\Longrightarrow\;s=t+2\;.$$

From $$k=-4-2t$$ (third equation) and the first equation, substitute $$s=t+2$$:

$$3t-(t+2)=2(-4-2t)\;.$$

Simplify:

$$2t-2=-8-4t\;\Longrightarrow\;6t=-6\;\Longrightarrow\;t=-1\;.$$

Hence $$s=t+2=1$$ and $$k=-4-2t=-2\;.$$

Thus the point on the first line is

$$P=(-2-3(-1),\;2+4(-1),\;5+2(-1))=(1,\,-2,\,3)\;,$$

and the required common perpendicular (shortest-distance) line is

$$\mathbf{r}=(1,\,-2,\,3)+\lambda\,(2,\,1,\,1)\;.$$

The point $$(-1,\alpha,\beta)$$ lies on this line, so

$$1+2\lambda=-1,\quad -2+\lambda=\alpha,\quad 3+\lambda=\beta\;.$$

From $$1+2\lambda=-1$$ we get $$\lambda=-1$$. Substituting:

$$\alpha=-2+(-1)=-3,\qquad \beta=3+(-1)=2\;.$$

Therefore

$$(\alpha-\beta)^2=(-3-2)^2=(-5)^2=25\;.$$

Hence, $$(\alpha-\beta)^2 = 25.$$

Line 1: $$(2+2s, -2s, 7+16s)$$. Line 2: $$(-3+4t, -2+3t, -2+t)$$.

Setting equal: $$2s - 4t = -5$$ ... (i), $$2s + 3t = 2$$ ... (ii), $$16s - t = -9$$ ... (iii).

Subtracting (i) from (ii): $$7t = 7$$, so $$t = 1$$. From (ii): $$s = -1/2$$. Check (iii): $$-8 - 1 = -9$$ \checkmark

Point $$P = (1, 1, -1)$$.

Distance from P to line $$\frac{x+1}{2} = \frac{y-1}{3} = \frac{z-1}{1}$$:

Point $$A(-1, 1, 1)$$, direction $$\vec{d} = (2, 3, 1)$$, $$\vec{AP} = (2, 0, -2)$$.

$$\vec{AP} \times \vec{d} = (6, -6, 6)$$, $$|\vec{AP} \times \vec{d}|^2 = 108$$.

$$l^2 = 108/14$$. $$14l^2 = \boxed{108}$$.

We need to find the square of the distance from the origin to the image of the point $$(6, 1, 5)$$ in the line $$\frac{x - 1}{3} = \frac{y}{2} = \frac{z - 2}{4}$$.

A general point on the line is:

$$P = (1 + 3t, \; 2t, \; 2 + 4t)$$

The direction from the given point $$(6, 1, 5)$$ to $$P$$ is:

$$(1 + 3t - 6, \; 2t - 1, \; 2 + 4t - 5) = (3t - 5, \; 2t - 1, \; 4t - 3)$$

For the perpendicular from the point to the line, this direction must be perpendicular to the line's direction vector $$(3, 2, 4)$$:

$$3(3t - 5) + 2(2t - 1) + 4(4t - 3) = 0$$

$$9t - 15 + 4t - 2 + 16t - 12 = 0$$

$$29t - 29 = 0 \implies t = 1$$

Substituting $$t = 1$$:

$$\text{Foot} = (1 + 3, \; 2, \; 2 + 4) = (4, 2, 6)$$

The image is the reflection of $$(6, 1, 5)$$ through the foot $$(4, 2, 6)$$. Since the foot is the midpoint of the original point and its image:

$$\text{Image} = (2 \times 4 - 6, \; 2 \times 2 - 1, \; 2 \times 6 - 5) = (2, 3, 7)$$

$$d^2 = 2^2 + 3^2 + 7^2 = 4 + 9 + 49 = 62$$

The answer is $$\textbf{62}$$.

The given plane is $$\sqrt{3}\,x+2y+3z=16$$.
For any point $$(\alpha,\beta,\gamma)$$ the (perpendicular) distance from this plane equals

$$\frac{\left|\sqrt{3}\alpha+2\beta+3\gamma-16\right|} {\sqrt{(\sqrt{3})^{2}+2^{2}+3^{2}}} \;=\;\frac{\left|\sqrt{3}\alpha+2\beta+3\gamma-16\right|}{4}.$$

We are told that this distance is $$\dfrac72$$, hence

$$\left|\sqrt{3}\alpha+2\beta+3\gamma-16\right|=14.$$

The left-hand side can never reach $$30$$ for a unit vector because
$$\max\bigl(\sqrt{3}\alpha+2\beta+3\gamma\bigr)=\sqrt{(\sqrt{3})^{2}+2^{2}+3^{2}}=4.$$
Therefore only the negative sign is possible and

$$\sqrt{3}\alpha+2\beta+3\gamma=2.$$ Thus $$S$$ is the circle obtained by the intersection of

• the unit sphere $$\alpha^{2}+\beta^{2}+\gamma^{2}=1,$$ and
• the plane $$\sqrt{3}x+2y+3z=2.$$

Let $$\vec{n}=(\sqrt{3},2,3),\;|\vec{n}|=4,\; \hat{n}=\dfrac{\vec{n}}{|\vec{n}|}.$$
The plane is at (signed) distance $$d=\frac{2}{4}=\frac12$$ from the origin along $$\hat{n}$$, so the circle has

centre $$\vec{C}=d\,\hat{n}=\frac12\hat{n},$$
radius $$r=\sqrt{1-d^{2}}=\sqrt{1-\frac14}=\frac{\sqrt3}{2}.$$

Choose two unit vectors $$\vec{e}_{1},\vec{e}_{2}$$ lying in the plane and orthogonal to each other such that $$\vec{e}_{1}\times\vec{e}_{2}=\hat{n}.$$

The three required vectors can now be written as vertices of an equilateral triangle of radius $$r$$ inside this circle:

$$\begin{aligned} \vec{u}&=\vec{C}+r\,\vec{e}_{1},\\ \vec{v}&=\vec{C}+r\Bigl(-\tfrac12\vec{e}_{1}+\tfrac{\sqrt3}{2}\vec{e}_{2}\Bigr),\\ \vec{w}&=\vec{C}+r\Bigl(-\tfrac12\vec{e}_{1}-\tfrac{\sqrt3}{2}\vec{e}_{2}\Bigr). \end{aligned}$$

Define
$$\vec{A}_{1}=\vec{e}_{1},\quad \vec{A}_{2}=-\tfrac12\vec{e}_{1}+\tfrac{\sqrt3}{2}\vec{e}_{2},\quad \vec{A}_{3}=-\tfrac12\vec{e}_{1}-\tfrac{\sqrt3}{2}\vec{e}_{2}.$$

Then $$\vec{u}=\vec{C}+r\vec{A}_{1},\; \vec{v}=\vec{C}+r\vec{A}_{2},\; \vec{w}=\vec{C}+r\vec{A}_{3}.$$

The volume of the parallelepiped is the absolute value of the scalar triple product

$$V=\left|\vec{u}\,\cdot\,(\vec{v}\times\vec{w})\right|.$$

Subtract the first column from the next two to simplify the determinant:

$$\bigl(\vec{u},\vec{v},\vec{w}\bigr)= \det\!\bigl[\vec{C}+r\vec{A}_{1},\; r(\vec{A}_{2}-\vec{A}_{1}),\; r(\vec{A}_{3}-\vec{A}_{1})\bigr]$$ $$=r^{2}\det\!\bigl[\vec{C}+r\vec{A}_{1},\; \vec{B}_{2},\;\vec{B}_{3}\bigr],$$ where $$\vec{B}_{2}=\vec{A}_{2}-\vec{A}_{1} =-\tfrac32\vec{e}_{1}+\tfrac{\sqrt3}{2}\vec{e}_{2},\quad \vec{B}_{3}=\vec{A}_{3}-\vec{A}_{1} =-\tfrac32\vec{e}_{1}-\tfrac{\sqrt3}{2}\vec{e}_{2}.$$

Because $$\vec{A}_{1},\vec{B}_{2},\vec{B}_{3}$$ all lie in the plane (they are perpendicular to $$\hat{n}$$),
$$\det\bigl[\vec{A}_{1},\vec{B}_{2},\vec{B}_{3}\bigr]=0,$$ so only $$\vec{C}$$ contributes:

$$\vec{u}\cdot(\vec{v}\times\vec{w})= r^{2}\det\!\bigl[\vec{C},\vec{B}_{2},\vec{B}_{3}\bigr] =r^{2}\,\vec{C}\cdot(\vec{B}_{2}\times\vec{B}_{3}).$$

The cross product is computed using $$\vec{e}_{1}\times\vec{e}_{2}=\hat{n}:$$ $$\vec{B}_{2}\times\vec{B}_{3} =\left(-\tfrac32\vec{e}_{1}+\tfrac{\sqrt3}{2}\vec{e}_{2}\right) \times \left(-\tfrac32\vec{e}_{1}-\tfrac{\sqrt3}{2}\vec{e}_{2}\right) =\frac{3\sqrt3}{2}\,\hat{n}.$$

Therefore

$$\begin{aligned} \vec{C}\cdot(\vec{B}_{2}\times\vec{B}_{3}) &=\frac12\hat{n}\cdot\frac{3\sqrt3}{2}\hat{n} =\frac{3\sqrt3}{4},\\[6pt] r^{2} &=\left(\frac{\sqrt3}{2}\right)^{2}=\frac34,\\[6pt] V &=\left|\frac34\cdot\frac{3\sqrt3}{4}\right| =\frac{9\sqrt3}{16}. \end{aligned}$$

Finally,

$$\frac{80}{\sqrt3}V =\frac{80}{\sqrt3}\cdot\frac{9\sqrt3}{16} =\frac{720}{16}=45.$$

Hence the required value is 45.

The two lines are
$$\ell_1: \;\vec r = \lambda(\hat i+\hat j+\hat k),\;\lambda\in\mathbb R$$ and
$$\ell_2: \;\vec r = (\hat j-\hat k)+\mu(\hat i+\hat k),\;\mu\in\mathbb R.$$

1. General form of a plane through $$\ell_1$$
Any plane $$H$$ containing $$\ell_1$$ must pass through the origin (because $$\ell_1$$ does) and must contain the direction vector $$\langle1,1,1\rangle$$.
If $$\mathbf n=\langle a,b,c\rangle$$ is the normal to $$H$$, then $$\mathbf n\cdot\langle1,1,1\rangle=0\;$$ i.e.
$$a+b+c=0\;-(1)$$
Hence every such plane has equation
$$ax+by+cz=0,\qquad a+b+c=0.$$\

2. Distance of a point on $$\ell_2$$ from $$H$$
A general point on $$\ell_2$$ is
$$P(\mu,\,1,\,-1+\mu).$$
Its perpendicular distance from $$H$$ is
$$d(\mu)=\frac{\lvert a\mu+b+c(-1+\mu)\rvert}{\sqrt{a^{2}+b^{2}+c^{2}}}$$ $$=\frac{\lvert\mu(a+c)+b-c\rvert}{\sqrt{a^{2}+b^{2}+c^{2}}}\;-(2).$$

3. Least distance from $$\ell_2$$ to $$H$$
For a fixed plane $$H$$ (fixed $$a,b,c$$) we can vary $$\mu$$ to minimise the numerator in (2).
• If $$a+c\neq0$$ we choose $$\mu=\dfrac{a+2c}{a+c}$$ to make the numerator zero, giving the minimum distance 0.
• If $$a+c=0$$, then from (1) $$b=-(a+c)=0$$ and $$c=-a$$.   The numerator in (2) becomes $$\lvert-a-2c\rvert=\lvert a\rvert$$, independent of $$\mu$$, so the least distance is
$$d(H)=\frac{\lvert a\rvert}{\sqrt{a^{2}+0+(-a)^{2}}}=\frac{\lvert a\rvert}{\lvert a\rvert\sqrt2}=\frac1{\sqrt2}\;-(3).$$

4. Plane $$H_0$$ giving the maximum of the least distances
For most planes the least distance is 0; the largest possible least distance (from (3)) is $$\dfrac1{\sqrt2}$$, attained when $$a+c=0$$.
Taking the simplest normal $$\mathbf n=\langle1,0,-1\rangle$$ (i.e. $$a=1,\,b=0,\,c=-1$$) we obtain
$$H_0:\;x-z=0\;(x=z).$$

Case P: $$d(H_0)=\dfrac1{\sqrt2}\Rightarrow(P)\to(5).$

5. Distance of (0,1,2) from $$H_0$$
$$d=$$\frac{\lvert0-2\rvert}{\sqrt{1^{2}$$+0^{2}+(-1)^{2}}}=$$\frac{2}{\sqrt2}=\sqrt$$2$$\Rightarrow$$(Q)$$\to$$(4).$$

6. Distance of origin from $$H_0$$
The origin lies on $$H_0$$, so the distance is 0 → $$(R)$$\to$$(3).$$

7. Point common to $$y=z,\;x=1,$$ and $$H_0$$
From $$x=1,\;x=z$$ we get $$z=1$$; from $$y=z$$, $$y=1$$. Thus the intersection point is $$A(1,1,1)$$.
Distance OA $$=$$\sqrt{1^{2}$$+1^{2}+1^{2}}=$$\sqrt$$3$$\Rightarrow$$(S)$$\to$$(1).$$

8. Final matching
$$$$\begin{aligned}$$ (P)&$$\to$$(5),\\ (Q)&$$\to$$(4),\\ (R)&$$\to$$(3),\\ (S)&$$\to$$(1). \end{aligned}$$
Hence the correct option is Option B.

The cube has side-length $$1$$ and its centre is at $$\left(\tfrac12,\tfrac12,\tfrac12\right)$$. Because of the full rotational symmetry of the cube, we may, without loss of generality, fix the body-diagonal

$$\ell_2:\; (0,0,0)+t(1,1,1),\;t\in\mathbb{R}$$

as the representative of the set $$S$$. For every face of the cube choose Cartesian coordinates in the natural way; each face diagonal has a direction vector which contains exactly two entries $$\pm1$$ and one entry $$0$$.

Write the direction vector of the fixed body diagonal as $$\mathbf{s}=(1,1,1)$$ and the direction vector of a face diagonal as $$\mathbf{f}=(a,b,c),\qquad (a,b,c)\in\{\! -1,0,1\!\},\;a^2+b^2+c^2=2,\;abc=0.$$ (The zero component indicates the coordinate which is constant on that face.)

Step 1 Magnitude of the cross product $$\mathbf{s}\times\mathbf{f}$$.
Since $$|\mathbf{s}|^2=3,\;|\mathbf{f}|^2=2$$, we use $$|\mathbf{s}\times\mathbf{f}|^2=|\mathbf{s}|^2|\mathbf{f}|^2-(\mathbf{s}\!\cdot\!\mathbf{f})^2=6-(\mathbf{s}\!\cdot\!\mathbf{f})^2.$$ Because $$\mathbf{s}\!\cdot\!\mathbf{f}=a+b+c,$$ the possible values are $$0,\;\pm2.$$ Hence

$$|\mathbf{s}\times\mathbf{f}|=\begin{cases}\sqrt6 & \text{if }\mathbf{s}\!\cdot\!\mathbf{f}=0,\\[4pt] \sqrt2 & \text{if }|\mathbf{s}\!\cdot\!\mathbf{f}|=2.\end{cases}$$

Step 2 Which face diagonals are skew to $$\ell_2$$?
If $$|\mathbf{s}\!\cdot\!\mathbf{f}|=2$$, the face diagonal lies in a face that contains the point $$(0,0,0)$$ or $$(1,1,1)$$, so it meets $$\ell_2$$. For intersecting lines the shortest distance is $$0$$, so such diagonals do not influence the maximum.
Therefore any face diagonal having a non-zero distance from $$\ell_2$$ must satisfy $$\mathbf{s}\!\cdot\!\mathbf{f}=0$$ and consequently $$|\mathbf{s}\times\mathbf{f}|=\sqrt6.$$

The vectors orthogonal to $$\mathbf{s}$$ with one zero entry are, up to sign and order,

$$(\;1,-1,0),\;(-1,1,0),\;(1,0,-1),\;(-1,0,1),\;(0,1,-1),\;(0,-1,1).$$

Each of these six vectors represents the two face diagonals on the two parallel faces whose normal is the coordinate axis corresponding to the zero entry. For definiteness consider

$$\ell_1:\;(1,0,1)+s(-1,1,0),\;s\in\mathbb{R}$$

which lies in the face $$z=1$$ and joins the vertices $$(1,0,1)$$ and $$(0,1,1).$$

Step 3 Distance between $$\ell_1$$ and $$\ell_2$$.
Take

$$P=(0,0,0)\in\ell_2,\qquad Q=(1,0,1)\in\ell_1,$$ $$\mathbf{u}=\mathbf{s}=(1,1,1),\qquad\mathbf{v}=\mathbf{f}=(-1,1,0).$$

The standard formula for the shortest distance between two skew lines is

$$d(\ell_1,\ell_2)=\frac{|(Q-P)\cdot(\mathbf{u}\times\mathbf{v})|}{|\mathbf{u}\times\mathbf{v}|}.$$

Compute the cross product: $$\mathbf{u}\times\mathbf{v} =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\[2pt] 1&1&1\\ -1&1&0 \end{vmatrix} =(-1,-1,2),\qquad|\mathbf{u}\times\mathbf{v}|=\sqrt6.$$

The triple product is $$(Q-P)\cdot(\mathbf{u}\times\mathbf{v}) =(1,0,1)\cdot(-1,-1,2)=(-1)+0+2=1.$$

Hence $$d(\ell_1,\ell_2)=\frac{1}{\sqrt6}.$$

Step 4 Showing this is the maximum.
For every face diagonal that is skew to $$\ell_2$$ the direction vector belongs to the above list, so $$|\mathbf{s}\times\mathbf{f}|=\sqrt6$$. Its two vertices are of the form $$(1,0,1)$$ & $$(0,1,1)$$ (or any permutation/sign-change of these coordinates). A direct substitution (as above) gives

$$|(Q-P)\cdot(\mathbf{s}\times\mathbf{f})|=1$$

for either choice of the end-point $$Q,$$ and therefore

$$d(\ell_1,\ell_2)=\frac{1}{\sqrt6}.$$

Lines with $$|\mathbf{s}\!\cdot\!\mathbf{f}|=2$$ meet $$\ell_2$$, giving distance $$0$$, so no larger value is possible. Since all four body diagonals and all twelve face diagonals are equivalent under cube symmetries, the same calculation applies to every pair in $$S\times F$$.

Thus the maximum distance between a line of $$F$$ and a line of $$S$$ is

$$\boxed{\dfrac{1}{\sqrt6}}$$

Option A is correct.

We are given point $$P(4, 6, -2)$$ and a line passing through $$A(-3, 2, 3)$$ with direction ratios $$(3, 3, -1)$$, so $$\vec{AP} = (4-(-3), 6-2, -2-3) = (7, 4, -5).$$

The direction vector of the line is $$\vec{d} = (3, 3, -1)$$, and the cross product of $$\vec{AP}$$ with $$\vec{d}$$ is given by:

$$\vec{AP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 4 & -5 \\ 3 & 3 & -1 \end{vmatrix} = \hat{i}(4 \cdot (-1) - (-5) \cdot 3) - \hat{j}(7 \cdot (-1) - (-5) \cdot 3) + \hat{k}(7 \cdot 3 - 4 \cdot 3)$$

$$= \hat{i}(-4 + 15) - \hat{j}(-7 + 15) + \hat{k}(21 - 12) = (11, -8, 9)$$

The magnitude of this cross product is $$|\vec{AP} \times \vec{d}| = \sqrt{121 + 64 + 81} = \sqrt{266}$$ while the magnitude of $$\vec{d}$$ is $$|\vec{d}| = \sqrt{9 + 9 + 1} = \sqrt{19}$$. It follows that the distance from the point to the line is $$\frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|} = \frac{\sqrt{266}}{\sqrt{19}} = \sqrt{\frac{266}{19}} = \sqrt{14}.$$

The correct answer is Option D: $$\sqrt{14}$$.

Consider the two lines $$L_1: \frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}$$ passing through the point $$A(1, -8, 4)$$ with direction vector $$\vec{d_1} = (2, -7, 5)$$ and $$L_2: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}$$ passing through the point $$B(1, 2, 6)$$ with direction vector $$\vec{d_2} = (2, 1, -3)$$.

The cross product of the direction vectors is $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21 - 5) - \hat{j}(-6 - 10) + \hat{k}(2 + 14) = (16, 16, 16),$$ and its magnitude is $$|\vec{d_1} \times \vec{d_2}| = 16\sqrt{3}.$$

The vector from A to B is $$\vec{AB} = B - A = (0, 10, 2).$$

The shortest distance between the lines is given by $$d = \frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} = \frac{|0\cdot16 + 10\cdot16 + 2\cdot16|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}.$$

Therefore, the correct answer is Option B: $$4\sqrt{3}$$.

Two vertices of triangle ABC are $$(2,4,6)$$ and $$(0,-2,-5)$$ with centroid $$(2,1,-1)$$. We need to find $$\alpha\beta + \beta\gamma + \gamma\alpha$$ where $$(\alpha,\beta,\gamma)$$ is the image of the third vertex in the plane $$x + 2y + 4z = 11$$.

Since the centroid formula gives $$G = \frac{A + B + C}{3}$$ and we have $$A = (2,4,6)$$, $$B = (0,-2,-5)$$, and $$G = (2,1,-1)$$, it follows that
$$x_C = 3(2) - 2 - 0 = 4$$
$$y_C = 3(1) - 4 - (-2) = 1$$
$$z_C = 3(-1) - 6 - (-5) = -4$$
Hence $$C = (4,1,-4)\,. $$

To reflect $$C$$ across the plane $$x + 2y + 4z = 11$$ (with parameters $$a=1,\,b=2,\,c=4,\,d=-11$$), we compute
$$t = \frac{-2\,(a x_0 + b y_0 + c z_0 + d)}{a^2 + b^2 + c^2} = \frac{-2\,(4 + 2 - 16 - 11)}{1 + 4 + 16} = \frac{-2(-21)}{21} = 2\,. $$
Therefore
$$\alpha = 4 + 1\cdot 2 = 6\,,\quad \beta = 1 + 2\cdot 2 = 5\,,\quad \gamma = -4 + 4\cdot 2 = 4\,. $$

Finally, the sum is
$$\alpha\beta + \beta\gamma + \gamma\alpha = 6\cdot5 + 5\cdot4 + 4\cdot6 = 30 + 20 + 24 = 74\,. $$

The line $$l_1$$ passes through the point $$A(2, 6, 2)$$ and is perpendicular to the plane $$2x + y - 2z = 10$$. Therefore, its direction vector $$\vec{d}_1$$ is equal to the normal vector of the plane:

$$\vec{d}_1 = 2\hat{i} + \hat{j} - 2\hat{k}$$

The second line $$l_2$$ is given by:

$$\frac{x+1}{2} = \frac{y+4}{-3} = \frac{z}{2}$$

From this equation, a point passing through $$l_2$$ is $$B(-1, -4, 0)$$ and its direction vector is:

$$\vec{d}_2 = 2\hat{i} - 3\hat{j} + 2\hat{k}$$

---

Step 1: Form the vector connecting the two points $$A$$ and $$B$$

The vector $$\vec{AB}$$ is calculated as:

$$\vec{AB} = (-1 - 2)\hat{i} + (-4 - 6)\hat{j} + (0 - 2)\hat{k} = -3\hat{i} - 10\hat{j} - 2\hat{k}$$

---

Step 2: Calculate the cross product of the direction vectors $$\vec{d}_1 \times \vec{d}_2$$

$$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix}$$

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

$$= -4\hat{i} - 8\hat{j} - 8\hat{k}$$

Finding the magnitude of this cross product vector:

$$\left|\vec{d}_1 \times \vec{d}_2\right| = \sqrt{(-4)^2 + (-8)^2 + (-8)^2} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12$$

---

Step 3: Evaluate the shortest distance using the scalar triple product

The shortest distance $$d$$ between two skew lines is given by the formula:

$$d = \frac{\left|\vec{AB} \cdot (\vec{d}_1 \times \vec{d}_2)\right|}{\left|\vec{d}_1 \times \vec{d}_2\right|}$$

Computing the dot product in the numerator:

$$\vec{AB} \cdot (\vec{d}_1 \times \vec{d}_2) = (-3)(-4) + (-10)(-8) + (-2)(-8)$$

$$\vec{AB} \cdot (\vec{d}_1 \times \vec{d}_2) = 12 + 80 + 16 = 108$$

Substituting the values into the shortest distance formula:

$$d = \frac{108}{12} = 9$$

Therefore, the shortest distance between the lines is equal to 9.

Consider the plane $$2x - y + z = 4$$ and the points $$A(a, -2, 4)$$ and $$B(2, b, -3)$$. The point $$C$$ divides the segment $$AB$$ internally in the ratio $$2:1$$.

The coordinates of $$C$$ are obtained by internal division in the ratio $$2:1$$: $$C = \left(\frac{2\cdot 2 + 1\cdot a}{3}, \frac{2\cdot b + 1\cdot(-2)}{3}, \frac{2\cdot(-3) + 1\cdot 4}{3}\right) = \left(\frac{4 + a}{3}, \frac{2b - 2}{3}, -\frac{2}{3}\right).$$

Since $$C$$ lies on the plane $$2x - y + z = 4$$, substitute its coordinates to get $$2\cdot\frac{4 + a}{3} - \frac{2b - 2}{3} + \left(-\frac{2}{3}\right) = 4.$$ Clearing denominators: $$8 + 2a - 2b + 2 - 2 = 12 \implies 8 + 2a - 2b = 12 \implies a - b = 2.$$

The condition $$|OC| = \sqrt{5}$$ implies $$|OC|^2 = 5$$. Computing the square of the distance from the origin to $$C$$ gives $$\frac{(4 + a)^2 + (2b - 2)^2 + (-2)^2}{9} = 5 \quad\Longrightarrow\quad (4 + a)^2 + (2b - 2)^2 = 41.$$

Using $$a = b + 2$$ from above and substituting into the last equation yields $$(6 + b)^2 + (2b - 2)^2 = 41,$$ which expands to $$36 + 12b + b^2 + 4b^2 - 8b + 4 = 41 \implies 5b^2 + 4b - 1 = 0 \implies b = \frac{-4 \pm 6}{10}.$$ Hence, $$b = \frac{1}{5}$$ or $$b = -1$$.

Since the product $$ab$$ must be negative, we select $$b = -1$$, giving $$a = 1$$ and $$ab = -1 < 0$$. The other value yields $$ab > 0$$.

Thus $$a = 1$$ and $$b = -1$$. The corresponding point is $$C = \left(\frac{5}{3}, -\frac{4}{3}, -\frac{2}{3}\right),$$ and the point $$P$$, defined by $$(a - b,\, b,\, 2b - a),$$ becomes $$P = (2, -1, -3).$$

Finally, the squared distance $$CP^2$$ is $$\left(2 - \frac{5}{3}\right)^2 + \left(-1 + \frac{4}{3}\right)^2 + \left(-3 + \frac{2}{3}\right)^2 = \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(-\frac{7}{3}\right)^2 = \frac{1 + 1 + 49}{9} = \frac{51}{9} = \frac{17}{3}.$$ The correct answer is Option A.

Find the intersection of lines $$\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{1}$$ and $$\frac{x-a}{2} = \frac{y+2}{3} = \frac{z-3}{1}$$. To this end, we introduce parameters $$t$$ and $$s$$ and write the first line as $$(x,y,z)=(1+t,2+2t,-3+t)$$ and the second line as $$(x,y,z)=(a+2s,-2+3s,3+s)$$.

Equating corresponding coordinates yields the system $$1+t = a + 2s,\quad 2+2t = -2+3s,\quad -3+t = 3+s.$$ From the third equation we get $$t = 6 + s$$. Substituting into the second equation gives $$2 + 2(6+s) = -2 + 3s\;\Longrightarrow\;14 + 2s = -2 + 3s\;\Longrightarrow\;s = 16,$$ and hence $$t = 22$$.

Substituting $$t = 22$$ into $$(1+t,2+2t,-3+t)$$ yields the intersection point $$P = (23,46,19).$$ Then the first equation $$1+t = a + 2s$$ becomes $$23 = a + 32\;\Longrightarrow\;a = -9.$$

To find the distance from $$P$$ to the plane $$z = a = -9$$, we compute $$|z_P - a| = |19 - (-9)| = 28.$$ Therefore, the correct answer is Option 2: $$28$$.

Let the three given points be $$A(1,2,3),\;B(2,3,4),\;C(1,5,7)$$. Any vector perpendicular to the plane $$ABC$$ is parallel to $$\overrightarrow{AB}\times\overrightarrow{AC}$$.

First write the two vectors lying in the plane: $$\overrightarrow{AB}=B-A=(1,1,1),\qquad \overrightarrow{AC}=C-A=(0,3,4).$$

Using the formula for a cross-product, $$\overrightarrow{AB}\times\overrightarrow{AC}= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 1\\ 0 & 3 & 4 \end{vmatrix} = \hat i(1\cdot4-1\cdot3)-\hat j(1\cdot4-1\cdot0)+\hat k(1\cdot3-1\cdot0) = 1\hat i-4\hat j+3\hat k.$$

Thus one normal vector is $$(1,-4,3)$$. Any scalar multiple of it is also normal to the plane. Because the required vector $$\vec{OP}$$ must be a unit vector and must satisfy $$\beta\in(0,\tfrac{\pi}{2})$$ (i.e. its y-component positive), we choose the opposite direction: $$(-1,4,-3).$$ This has a positive y-component ($$+4$$), so $$\beta$$ indeed lies between $$0$$ and $$\tfrac{\pi}{2}$$.

Convert this direction to a unit vector: $$\sqrt{(-1)^2+4^2+(-3)^2}=\sqrt{1+16+9}=\sqrt{26},$$ so $$\vec{OP}= \frac{-1}{\sqrt{26}}\hat i+\frac{4}{\sqrt{26}}\hat j+\frac{-3}{\sqrt{26}}\hat k.$$

Therefore $$\cos\alpha=\frac{-1}{\sqrt{26}}\lt0,\qquad \cos\beta=\frac{4}{\sqrt{26}}\gt0,\qquad \cos\gamma=\frac{-3}{\sqrt{26}}\lt0.$$

When the cosine is negative, the angle lies in $$(\tfrac{\pi}{2},\pi)$$; when the cosine is positive, the angle lies in $$(0,\tfrac{\pi}{2})$$. Hence $$\alpha\in\left(\tfrac{\pi}{2},\pi\right),\qquad \beta\in\left(0,\tfrac{\pi}{2}\right),\qquad \gamma\in\left(\tfrac{\pi}{2},\pi\right).$$

Comparing with the given options, this matches Option A: $$\alpha \in (\tfrac{\pi}{2},\pi)\ \text{and}\ \gamma \in (\tfrac{\pi}{2},\pi).$$

To find the image of $$P(2, 3, 5)$$ in the plane $$2x + y - 3z = 6$$.

The formula for the image $$(\alpha, \beta, \gamma)$$ of point $$(x_1, y_1, z_1)$$ in plane $$ax + by + cz = d$$ is:

$$\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 - d)}{a^2 + b^2 + c^2}$$

$$\frac{-2(2(2) + 1(3) - 3(5) - 6)}{4 + 1 + 9} = \frac{-2(4 + 3 - 15 - 6)}{14} = \frac{-2(-14)}{14} = 2$$

Therefore:

$$\alpha = 2 + 2(2) = 6$$

$$\beta = 3 + 2(1) = 5$$

$$\gamma = 5 + 2(-3) = -1$$

$$\alpha + \beta + \gamma = 6 + 5 + (-1) = 10$$

$$\overrightarrow{AB} = (9 - (-1))\hat{i} + (3 - (-2))\hat{j} + (4 - (-3))\hat{k} = 10\hat{i} + 5\hat{j} + 7\hat{k}$$

$$\overrightarrow{AC} = (9 - (-1))\hat{i} + (-2 - (-2))\hat{j} + (1 - (-3))\hat{k} = 10\hat{i} + 0\hat{j} + 4\hat{k}$$

$$\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \text{det} \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 10 & 5 & 7 \\ 10 & 0 & 4 \end{bmatrix}$$

$$\vec{n} = \hat{i}(5 \cdot 4 - 7 \cdot 0) - \hat{j}(10 \cdot 4 - 7 \cdot 10) + \hat{k}(10 \cdot 0 - 5 \cdot 10)$$

$$\vec{n} = 20\hat{i} - \hat{j}(40 - 70) - 50\hat{k} = 20\hat{i} + 30\hat{j} - 50\hat{k}$$

$$\vec{n}_{\text{scaled}} = 2\hat{i} + 3\hat{j} - 5\hat{k}$$

Using point $$C(9, -2, 1)$$, the scalar equation of the plane is:

$$2(x - 9) + 3(y - (-2)) - 5(z - 1) = 0$$

$$2x - 18 + 3y + 6 - 5z + 5 = 0$$

$$2x + 3y - 5z - 7 = 0$$

Using the foot of the perpendicular formula:

$$\frac{\alpha - 3}{2} = \frac{\beta - (-2)}{3} = \frac{\gamma - (-9)}{-5} = -\frac{2(3) + 3(-2) - 5(-9) - 7}{2^2 + 3^2 + (-5)^2}$$

$$\text{Ratio} = -\frac{38}{38} = -1$$

$$\frac{\alpha - 3}{2} = -1 \implies \alpha - 3 = -2 \implies \alpha = 1$$

$$\frac{\beta + 2}{3} = -1 \implies \beta + 2 = -3 \implies \beta = -5$$

$$\frac{\gamma + 9}{-5} = -1 \implies \gamma + 9 = 5 \implies \gamma = -4$$

The coordinates of the foot of the perpendicular are $$Q(1, -5, -4)$$

$$OQ = \sqrt{1^2 + (-5)^2 + (-4)^2}$$

$$OQ = \sqrt{1 + 25 + 16} = \sqrt{42}$$

We seek the reflection $$Q(\alpha,\beta,\gamma)$$ of the point $$P(1,2,6)$$ in the plane determined by $$A(1,2,0)$$, $$B(1,4,1)$$ and $$C(0,5,1)$$, and then compute $$\alpha^2+\beta^2+\gamma^2$$.

First, the vectors in the plane are

A normal vector follows from their cross product:

Using $$A(1,2,0)$$, the plane equation becomes

To reflect a general point $$(x_1,y_1,z_1)$$ in the plane $$ax+by+cz+d=0$$ one uses

Here $$a=1,\;b=1,\;c=-2,\;d=-3$$ and for $$P(1,2,6)$$ we compute

Hence

giving

Finally,

The correct answer is Option 1: 65.

We need to find the image of point $$P(2, -1, 3)$$ in the plane $$x + 2y - z = 0$$, and then find its distance from another plane.

Recall the formula for the image of a point in a plane.

The image $$Q$$ of a point $$P(x_0, y_0, z_0)$$ in the plane $$ax + by + cz + d = 0$$ is found using the formula:

$$ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = \frac{-2(ax_0 + by_0 + cz_0 + d)}{a^2 + b^2 + c^2} $$

This formula works because the line from $$P$$ to $$Q$$ is perpendicular to the plane (direction ratios $$(a, b, c)$$), and the midpoint of $$PQ$$ lies on the plane.

Apply the formula with the given values.

Plane: $$x + 2y - z = 0$$, so $$a = 1, b = 2, c = -1, d = 0$$.

Point: $$P(2, -1, 3)$$.

Calculate $$ax_0 + by_0 + cz_0 + d$$:

$$ 1(2) + 2(-1) + (-1)(3) + 0 = 2 - 2 - 3 = -3 $$

Calculate $$a^2 + b^2 + c^2$$:

$$ 1 + 4 + 1 = 6 $$

The parameter value:

$$ t = \frac{-2(-3)}{6} = \frac{6}{6} = 1 $$

Find the coordinates of the image $$Q$$.

$$ x = x_0 + at = 2 + 1(1) = 3 $$

$$ y = y_0 + bt = -1 + 2(1) = 1 $$

$$ z = z_0 + ct = 3 + (-1)(1) = 2 $$

So $$Q = (3, 1, 2)$$.

Verify that $$Q$$ is the correct image.

The midpoint of $$PQ$$ is $$\left(\frac{2+3}{2}, \frac{-1+1}{2}, \frac{3+2}{2}\right) = (2.5, 0, 2.5)$$.

Check if it lies on the plane: $$2.5 + 2(0) - 2.5 = 0$$ . Confirmed.

Find the distance from $$Q(3, 1, 2)$$ to the plane $$3x + 2y + z + 29 = 0$$.

The distance from a point $$(x_0, y_0, z_0)$$ to a plane $$ax + by + cz + d = 0$$ is:

$$ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} $$

$$ D = \frac{|3(3) + 2(1) + 1(2) + 29|}{\sqrt{9 + 4 + 1}} = \frac{|9 + 2 + 2 + 29|}{\sqrt{14}} = \frac{42}{\sqrt{14}} $$

Rationalizing:

$$ D = \frac{42}{\sqrt{14}} = \frac{42\sqrt{14}}{14} = 3\sqrt{14} $$

The distance is $$3\sqrt{14}$$.

The correct answer is Option 4: $$3\sqrt{14}$$.

The line $$L$$ passes through $$(0, 1, 2)$$, intersects the line $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$, and is parallel to the plane $$2x + y - 3z = 4$$.

Find the intersection point.

A general point on the given line is $$(1 + 2\lambda, 2 + 3\lambda, 3 + 4\lambda)$$.

Direction ratios of $$L$$ from $$(0, 1, 2)$$ to this point: $$(1 + 2\lambda,\ 1 + 3\lambda,\ 1 + 4\lambda)$$.

Apply the parallel-to-plane condition.

Since $$L$$ is parallel to the plane $$2x + y - 3z = 4$$, its direction is perpendicular to the normal $$(2, 1, -3)$$:

$$2(1 + 2\lambda) + (1 + 3\lambda) - 3(1 + 4\lambda) = 0$$

$$2 + 4\lambda + 1 + 3\lambda - 3 - 12\lambda = 0 \implies -5\lambda = 0 \implies \lambda = 0$$

Determine line $$L$$.

The intersection point is $$(1, 2, 3)$$, so $$L$$ passes through $$(0, 1, 2)$$ and $$(1, 2, 3)$$ with direction $$(1, 1, 1)$$.

Distance from $$P(1, -9, 2)$$ to line $$L$$.

$$\vec{AP} = P - A = (1 - 0,\ -9 - 1,\ 2 - 2) = (1, -10, 0)$$

$$\vec{d} = (1, 1, 1)$$

$$\vec{AP} \times \vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -10 & 0 \\ 1 & 1 & 1 \end{vmatrix} = (-10 - 0)\vec{i} - (1 - 0)\vec{j} + (1 + 10)\vec{k} = (-10, -1, 11)$$

$$|\vec{AP} \times \vec{d}|^2 = 100 + 1 + 121 = 222$$

$$|\vec{d}|^2 = 1 + 1 + 1 = 3$$

$$\text{Distance} = \frac{\sqrt{222}}{\sqrt{3}} = \sqrt{\frac{222}{3}} = \sqrt{74}$$

The answer is Option A: $$\sqrt{74}$$.

Given lines $$L_1: \frac{x+5}{3} = \frac{y+4}{1} = \frac{z-\alpha}{-2}$$ and $$L_2$$ is the intersection of planes $$3x + 2y + z = 2$$ and $$x - 3y + 2z = 13$$.

Find direction and point of $$L_2$$.

Direction of $$L_2$$ = $$\vec{n_1} \times \vec{n_2} = (3,2,1) \times (1,-3,2) = (7,-5,-11)$$

A point on $$L_2$$: Setting $$z = 0$$: $$3x + 2y = 2$$ and $$x - 3y = 13$$.

Solving: $$x = \frac{32}{11}$$, $$y = -\frac{37}{11}$$. Point $$B = \left(\frac{32}{11}, -\frac{37}{11}, 0\right)$$.

Find $$\alpha$$ using coplanarity condition.

Point on $$L_1$$: $$A = (-5, -4, \alpha)$$, direction $$\vec{d_1} = (3, 1, -2)$$.

$$\vec{d_1} \times \vec{d_2} = (3,1,-2) \times (7,-5,-11) = (-21, 19, -22)$$

$$\vec{AB} = \left(-\frac{87}{11}, -\frac{7}{11}, \alpha\right) \cdot (-21, 19, -22) = 0$$

$$\frac{87 \times 21}{11} - \frac{7 \times 19}{11} - 22\alpha = 0$$

$$\frac{1827 - 133}{11} = 22\alpha \implies \frac{1694}{11} = 22\alpha \implies \alpha = 7$$

Find point P on $$L_1$$ nearest to $$Q(-4, -3, 2)$$.

Parametric form of $$L_1$$: $$P = (-5+3t, -4+t, 7-2t)$$.

$$\vec{PQ} = (-4-(-5+3t), -3-(-4+t), 2-(7-2t)) = (1-3t, 1-t, -5+2t)$$

For nearest point, $$\vec{PQ} \cdot \vec{d_1} = 0$$:

$$3(1-3t) + 1(1-t) + (-2)(-5+2t) = 0$$

$$3 - 9t + 1 - t + 10 - 4t = 0 \implies 14 - 14t = 0 \implies t = 1$$

$$P = (-5+3, -4+1, 7-2) = (-2, -3, 5)$$

$$|a| + |b| + |c| = |-2| + |-3| + |5| = 2 + 3 + 5 = 10$$

Therefore, the correct answer is Option D: $$\mathbf{10}$$.

The rectangular parallelepiped has vertices at the origin O(0,0,0) and edges along the axes with lengths 3, 4, 5. The vertex P is at (3,4,5).

The diagonal OP has direction ratios (3,4,5) and passes through O(0,0,0).

Edges parallel to the z-axis not passing through O or P are at vertices (3,0,z), (0,4,z), and (3,4,z). But (3,4,z) passes through P at z=5, so the relevant edges not through O or P are at (3,0,z) and (0,4,z).

For edge at (3,0,z): The line is $$\vec{r_1} = (3,0,0) + s(0,0,1)$$.

The line OP is $$\vec{r_2} = t(3,4,5)$$.

The shortest distance formula between two skew lines is:

$$ d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} $$

Here $$\vec{a_1} = (0,0,0)$$, $$\vec{a_2} = (3,0,0)$$, $$\vec{d_1} = (3,4,5)$$, $$\vec{d_2} = (0,0,1)$$.

$$ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 0 & 0 & 1 \end{vmatrix} = (4)\hat{i} - (3)\hat{j} + (0)\hat{k} = (4, -3, 0) $$

$$|\vec{d_1} \times \vec{d_2}| = \sqrt{16+9} = 5$$

$$(\vec{a_2}-\vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = (3,0,0) \cdot (4,-3,0) = 12$$

$$ d = \frac{|12|}{5} = \frac{12}{5} $$

The shortest distance is $$\dfrac{12}{5}$$.

We need to find the distance of $$(7, -3, -4)$$ from the plane containing $$(2, -3, 1)$$, $$(-1, 1, -2)$$, and $$(3, -4, 2)$$.

We start by finding two vectors in the plane, namely $$\vec{AB} = (-3, 4, -3)$$ and $$\vec{AC} = (1, -1, 1)$$, where $$\vec{AB} = (-1-2, 1+3, -2-1)$$ and $$\vec{AC} = (3-2, -4+3, 2-1)$$.

Next, we determine a normal vector to the plane by computing the cross product $$\vec{n} = \vec{AB} \times \vec{AC}$$: $$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 4 & -3 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-3+3) + \hat{k}(3-4) = (1, 0, -1).$$

Using the point $$(2, -3, 1)$$ and the normal vector $$(1, 0, -1)$$, the plane equation is $$1(x-2) + 0(y+3) - 1(z-1) = 0$$ which simplifies to $$x - 2 - z + 1 = 0$$ and hence $$x - z - 1 = 0.$$

Finally, the distance from $$(7, -3, -4)$$ to the plane $$x - z - 1 = 0$$ is given by $$d = \frac{|7 - (-4) - 1|}{\sqrt{1 + 0 + 1}} = \frac{|7 + 4 - 1|}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}.$$

Therefore, the required distance is $$5\sqrt{2}$$.