Let $$\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$, $$\vec{b} = \hat{i} - 2\hat{j} - 2\hat{k}$$ and $$\vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k}$$. If $$\vec{d}$$ is a vector perpendicular to both $$\vec{b}$$ and $$\vec{c}$$, and $$\vec{a} \cdot \vec{d} = 18$$, then $$|\vec{a} \times \vec{d}|^2$$ is equal to

Given: $$\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$, $$\vec{b} = \hat{i} - 2\hat{j} - 2\hat{k}$$, $$\vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k}$$.

Since $$\vec{d}$$ is perpendicular to both $$\vec{b}$$ and $$\vec{c}$$, $$\vec{d}$$ is parallel to $$\vec{b} \times \vec{c}$$.

$$ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -2 \\ -1 & 4 & 3 \end{vmatrix} = \hat{i}(-6+8) - \hat{j}(3-2) + \hat{k}(4-2) = 2\hat{i} - \hat{j} + 2\hat{k} $$

Let $$\vec{d} = t(2\hat{i} - \hat{j} + 2\hat{k})$$.

Using $$\vec{a} \cdot \vec{d} = 18$$:

$$ t(2 \cdot 2 + 3 \cdot (-1) + 4 \cdot 2) = t(4-3+8) = 9t = 18 \implies t = 2 $$

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

$$ \vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & -2 & 4 \end{vmatrix} = \hat{i}(12+8) - \hat{j}(8-16) + \hat{k}(-4-12) = 20\hat{i} + 8\hat{j} - 16\hat{k} $$

$$ |\vec{a} \times \vec{d}|^2 = 400 + 64 + 256 = 720 $$

The correct answer is 720.