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

Write the vector $$\vec{a}$$ in component form as $$\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$$.

First find each cross-product.

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

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

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

Next take the square (magnitude squared) of each vector:

$$\bigl|\vec{a}\times\hat{i}\bigr|^{2}=0^{2}+a_3^{2}+a_2^{2}=a_2^{2}+a_3^{2}$$

$$\bigl|\vec{a}\times\hat{j}\bigr|^{2}=(-a_3)^{2}+0^{2}+a_1^{2}=a_3^{2}+a_1^{2}$$

$$\bigl|\vec{a}\times\hat{k}\bigr|^{2}=a_2^{2}+(-a_1)^{2}+0^{2}=a_2^{2}+a_1^{2}$$

Add the three results:

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

But $$a_1^{2}+a_2^{2}+a_3^{2}=|\vec{a}|^{2}=\vec{a}^{2}$$, so

$$(\vec{a}\times\hat{i})^{2}+(\vec{a}\times\hat{j})^{2}+(\vec{a}\times\hat{k})^{2}=2\vec{a}^{2}$$

Therefore the required value equals $$2\vec{a}^{2}$$.

Option C which is: $$2\vec{a}^2$$