If $\vec{a}$ is any vector, then $|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2$ is equal to

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

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

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

Using $\hat{i} \times \hat{i} = 0$, $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{k} \times \hat{i} = \hat{j}$:

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

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

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

Using $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{j} = 0$, $\hat{k} \times \hat{j} = -\hat{i}$:

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

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

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

Using $\hat{i} \times \hat{k} = -\hat{j}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{k} = 0$:

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

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

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Adding the three magnitudes:

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

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

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

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

$= 2|\vec{a}|^2$

Therefore, the answer is $2|\vec{a}|^2$.