For any vector $\vec{a}$, the value of $(\vec{a} \times \hat{i})^2 + (\vec{a} \times \hat{j})^2 + (\vec{a} \times \hat{k})^2$ is equal to:

Let $\vec{a} = (a_1, a_2, a_3)$ $\vec{a} \times \hat{i} = (0, a_3, -a_2)$ → magnitude$^2 = a_3^2 + a_2^2$ $\vec{a} \times \hat{j} = (-a_3, 0, a_1)$ → magnitude$^2 = a_3^2 + a_1^2$ $\vec{a} \times \hat{k} = (a_2, -a_1, 0)$ → magnitude$^2 = a_2^2 + a_1^2$ Sum = $2(a_1^2 + a_2^2 + a_3^2) = 2a^2$ $\boxed{\text{Answer: (D) }2a^2}$