Magnitude in Component Form

$$M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}$$

$$\vec{u} = \alpha\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \gamma\hat{k}$$

$$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \beta\hat{j} + \delta\hat{k}$$

$$\vec{w} = -\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \mu\hat{k}$$

$$\alpha^2 + \frac{1}{3} + \gamma^2 = 1 \quad \therefore \alpha^2 + \gamma^2 = \frac{2}{3} \quad \dots(1)$$

$$\frac{1}{2} + \beta^2 + \delta^2 = 1 \quad \beta^2 + \delta^2 = \frac{1}{2} \quad \dots(2)$$

$$\frac{1}{2} + \frac{1}{3} + \mu^2 = 1 \quad \mu^2 = \frac{1}{6} \quad \dots(3)$$

$$\gamma^2 + \delta^2 + \mu^2 = 1$$

$$\gamma^2 + \delta^2 = 1 - \frac{1}{6} = \frac{5}{6}$$

For Q:

$$x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}$$

$$x\vec{u} \cdot \vec{u} + y\vec{v} \cdot \vec{u} + z\vec{w} \cdot \vec{u} = \vec{u} \cdot \hat{j}$$

$$x = \frac{1}{\sqrt{3}}$$

For R: 

$$\vec{u}, \vec{v}, \vec{w}$$ are mutually perpendicular unit vectors.

$$\vert{}\vec{u} \cdot (\vec{v} \times \vec{w})\vert{}$$

$$\begin{bmatrix} \vec{u} & \vec{v} & \vec{w} \end{bmatrix} = 1$$

For S: 

$$\vec{u} \times (\vec{v} \times \vec{w})$$

$$(\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$$

$$= 0 - 0 = 0$$

Since $$\vec{d}$$ is coplanar with $$\vec{a}$$ and $$\vec{b}$$, it can be written as a linear combination of them:

$$\vec{d} = \lambda \vec{a} + \mu \vec{b}$$

Substitute the given vectors $$\vec{a}$$ and $$\vec{b}$$:

$$\vec{d} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})$$ $$\vec{d} = (\lambda + \mu)\hat{i} + \lambda - \mu)\hat{j} + (\lambda + \mu)\hat{k}$$

We are given that $$\vec{d}$$ is perpendicular to $$\vec{a}$$ ($$\vec{d} \cdot \vec{a} = 0$$):

$$(\lambda \vec{a} + \mu \vec{b}) \cdot \vec{a} = 0$$ $$\lambda |\vec{a}|^2 + \mu (\vec{b} \cdot \vec{a}) = 0$$

Let's calculate the required dot products:

• $$|\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3$$
• $$\vec{b} \cdot \vec{a} = (1)(1) + (-1)(1) + (1)(1) = 1 - 1 + 1 = 1$$

Substitute these values back into the condition:

$$3\lambda + \mu(1) = 0 \implies \mu = -3\lambda$$

Now, update our expression for $$\vec{d}$$ in terms of $$\lambda$$ only:

$$\vec{d} = \lambda \vec{a} - 3\lambda \vec{b} = \lambda (\vec{a} - 3\vec{b})$$

Let's compute $$\vec{a} - 3\vec{b}$$:

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

So, the vector $$\vec{d}$$ simplifies to:

$$\vec{d} = \lambda(-2\hat{i} + 4\hat{j} - 2\hat{k})$$

We are given $$\vec{d} \cdot \vec{c} = 6$$, where $$\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$$:

$$\lambda(-2\hat{i} + 4\hat{j} - 2\hat{k}) \cdot (\hat{i} + 2\hat{j} - \hat{k}) = 6$$ $$\lambda \left[ (-2)(1) + (4)(2) + (-2)(-1) \right] = 6$$ $$\lambda \left[ -2 + 8 + 2 \right] = 6$$ $$8\lambda = 6 \implies \lambda = \frac{3}{4}$$

Now we can write the exact vector $$\vec{d}$$:

$$\vec{d} = \frac{3}{4}(-2\hat{i} + 4\hat{j} - 2\hat{k}) = -\frac{3}{2}\hat{i} + 3\hat{j} - \frac{3}{2}\hat{k}$$

Finally, compute the square of its magnitude:

$$|\vec{d}|^2 = \left(-\frac{3}{2}\right)^2 + (3)^2 + \left(-\frac{3}{2}\right)^2$$ $$|\vec{d}|^2 = \frac{9}{4} + 9 + \frac{9}{4}$$ $$|\vec{d}|^2 = \frac{18}{4} + 9 = \frac{9}{2} + 9 = \frac{27}{2}$$

Correct Answer: C. $$\frac{27}{2}$$

Let's first simplify the inner cross product vector term, $$\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} \times \vec{b}\right)$$, by distributing it:

$$\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} \times \vec{b}\right) = \vec{a} \times \left(\vec{a} \times \vec{b}\right) + \vec{b} \times \left(\vec{a} \times \vec{b}\right)$$

Using the vector triple product expansion formula $$\vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$$:

1. $$\vec{a} \times \left(\vec{a} \times \vec{b}\right) = (\vec{a} \cdot \vec{b})\vec{a} - |\vec{a}|^2\vec{b}$$
2. $$\vec{b} \times \left(\vec{a} \times \vec{b}\right) = -\vec{b} \times \left(\vec{b} \times \vec{a}\right) = -\left[ (\vec{b} \cdot \vec{a})\vec{b} - |\vec{b}|^2\vec{a} \right] = |\vec{b}|^2\vec{a} - (\vec{a} \cdot \vec{b})\vec{b}$$

Adding these two results together:

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

Now, we compute the final outer cross product with $(\vec{a} - \vec{b})$:

$$\left[ \left( \vec{a} \cdot \vec{b} + |\vec{b}|^2 \right)\vec{a} - \left( |\vec{a}|^2 + \vec{a} \cdot \vec{b} \right)\vec{b} \right] \times (\vec{a} - \vec{b})$$

Since $$\vec{a} \times \vec{a} = 0$$ and $$\vec{b} \times \vec{b} = 0$$, this distributes cleanly into:

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

Step 2: Determine the value of $$\lambda$$

Now let's compute $$|\vec{a}|^2$$ and $$|\vec{b}|^2$$ from our given vectors:

• $$|\vec{a}|^2 = \lambda^2 + 3^2 + (-1)^2 = \lambda^2 + 10$$
• $$|\vec{b}|^2 = 2^2 + \lambda^2 + 1^2 = \lambda^2 + 5$$

Subtracting them gives a constant:

$$|\vec{a}|^2 - |\vec{b}|^2 = (\lambda^2 + 10) - (\lambda^2 + 5) = 5$$

Substitute this back into our simplified relation:

$$5(\vec{a} \times \vec{b}) = 25\hat{i} - 20\hat{j} - 10\hat{k}$$ $$\vec{a} \times \vec{b} = 5\hat{i} - 4\hat{j} - 2\hat{k}$$

Let's verify this using the cross-product determinant of $$\vec{a}$$ and $$\vec{b}$$:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda & 3 & -1 \\ 2 & \lambda & 1 \end{vmatrix} = (3 + \lambda)\hat{i} - (\lambda + 2)\hat{j} + (\lambda^2 - 6)\hat{k}$$

Comparing components with $$5\hat{i} - 4\hat{j} - 2\hat{k}$$:

• $$3 + \lambda = 5 \implies \lambda = 2$$
• $$-(\lambda + 2) = -4 \implies \lambda = 2$$

Both yield $$\lambda = 2$$.

Step 3: Find the target expression value

We need to find $$\left| \lambda\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} - \vec{b}\right) \right|^2$$.

Expanding the cross product inside the norm:

$$\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} - \vec{b}\right) = \vec{a}\times\vec{a} - \vec{a}\times\vec{b} + \vec{b}\times\vec{a} - \vec{b}\times\vec{b} = -2(\vec{a} \times \vec{b})$$

Substitute this into the expression:

$$\left| -2\lambda (\vec{a} \times \vec{b}) \right|^2 = 4\lambda^2 |\vec{a} \times \vec{b}|^2$$

Using $$\lambda = 2$$ and $$|\vec{a} \times \vec{b}|^2 = 5^2 + (-4)^2 + (-2)^2 = 25 + 16 + 4 = 45$$:

$$\text{Value} = 4(2)^2 \cdot 45 = 16 \cdot 45 = 720$$

Correct Option: (3) 720