If $|\vec a| = a$, then the value of $|\vec a × \hat i|^2 + |\vec a × \hat j|^2 + |\vec a × \vec k|^2$ is

$2a^2$

The correct answer is Option (2) → $2a^2$

Let $\vec{a}$ be a vector such that $|\vec{a}| = a$ and $\hat{i}, \hat{j}, \hat{k}$ be unit vectors along x, y, z axes.

We know: $|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = |\vec{a}|^2 + |\vec{a}|^2 + |\vec{a}|^2 - 2(a_x^2 + a_y^2 + a_z^2)$

Or more directly, using $\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$:

$|\vec{a} \times \hat{i}|^2 = a_y^2 + a_z^2$

$|\vec{a} \times \hat{j}|^2 = a_x^2 + a_z^2$

$|\vec{a} \times \hat{k}|^2 = a_x^2 + a_y^2$

Add them: $(a_y^2 + a_z^2) + (a_x^2 + a_z^2) + (a_x^2 + a_y^2) = 2(a_x^2 + a_y^2 + a_z^2) = 2|\vec{a}|^2 = 2a^2$