For any vector $\vec a$, the value of $|\vec a×\hat i|^2 + |\vec a×\hat j|^2 + |\vec a×\hat k|^2$ is equal to:

$2|\vec a|^2$

The correct answer is Option (1) → $2|\vec a|^2$

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$

Compute each cross product:

$\vec{a} \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 1 & 0 & 0 \end{vmatrix} = (0 - 0)\hat{i} - (0 - a_3)\hat{j} + (0 - a_2)\hat{k} = a_3\hat{j} - a_2\hat{k}$

$|\vec{a} \times \hat{i}|^2 = a_3^2 + a_2^2$

Similarly,

$\vec{a} \times \hat{j} = -a_3\hat{i} + a_1\hat{k}$

$|\vec{a} \times \hat{j}|^2 = a_3^2 + a_1^2$

$\vec{a} \times \hat{k} = a_2\hat{i} - a_1\hat{j}$

$|\vec{a} \times \hat{k}|^2 = a_2^2 + a_1^2$

Now add them:

$|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2$

$= (a_2^2 + a_3^2) + (a_3^2 + a_1^2) + (a_1^2 + a_2^2)$

$= 2(a_1^2 + a_2^2 + a_3^2)$

$= 2|\vec{a}|^2$

Therefore, the value is $2|\vec{a}|^2$.