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:

Correct option is C

Given:
$$\vec{a}$$​ is any vector.
$$\hat{i}, \hat{j}, \hat{k}$$​ are unit vectors along x, y, z axes.
Formula used:
​$$|\vec{a} \times \hat{n}|^{2} = |\vec{a}|^{2} - (\vec{a}\cdot \hat{n})^{2}$$​​
Solution:
Let
​$$\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$$​​
Then,
​$$(\vec{a} \times \hat{i})^{2}= \vec{a}^{\,2} - a_x^{2}\\(\vec{a} \times \hat{j})^{2}= \vec{a}^{\,2} - a_y^{2}\\(\vec{a} \times \hat{k})^{2}= \vec{a}^{\,2} - a_z^{2}$$​​
Adding:
​$$(\vec{a} \times \hat{i})^{2}+ (\vec{a} \times \hat{j})^{2}+ (\vec{a} \times \hat{k})^{2}= 3\vec{a}^{\,2} - (a_x^{2} + a_y^{2} + a_z^{2})$$​​
But,
​$$a_x^{2} + a_y^{2} + a_z^{2} = \vec{a}^{\,2}$$​​
Hence,
​$$= 3\vec{a}^{\,2} - \vec{a}^{\,2}= 2\vec{a}^{\,2}$$​​
The correct answer is (c) $$2\vec{a}^{\,2}.$$​​