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

Given $|\vec{a}| = a$, where $a$ is the magnitude of vector $\vec{a}$.

Any vector can be written as:

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

where $a_1, a_2, a_3$ are the components along x, y, and z axes.

Since $|\vec{a}| = a$:

$a_1^2 + a_2^2 + a_3^2 = a^2$

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For $\vec{a} \times \hat{i}$:

$\vec{a} \times \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{i}$

$\vec{a} \times \hat{i} = a_3\hat{j} - a_2\hat{k}$

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

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For $\vec{a} \times \hat{j}$:

$\vec{a} \times \hat{j} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{j}$

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

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

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For $\vec{a} \times \hat{k}$:

$\vec{a} \times \hat{k} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{k}$

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

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

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

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

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

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

$= 2a^2$

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