1. The problem states: If \(\vec{A}\) is a non-zero vector, we want to find the value of: $$\|\vec{A} \times \hat{i}\|^2 + \|\vec{A} \times \hat{j}\|^2 + \|\vec{A} \times \hat{k}\|^2$$ where \(\hat{i}, \hat{j}, \hat{k}\) are the standard unit vectors in 3D. 2. Recall that for any vectors \(\vec{u}, \vec{v}\), the magnitude of their cross product squared is: $$\|\vec{u} \times \vec{v}\|^2 = \|\vec{u}\|^2 \|\vec{v}\|^2 - (\vec{u} \cdot \vec{v})^2$$ 3. Since \(\hat{i}, \hat{j}, \hat{k}\) are unit vectors, \(\|\hat{i}\| = \|\hat{j}\| = \|\hat{k}\| = 1\). 4. Let \(\vec{A} = (a_1, a_2, a_3)\). Then: - \(\|\vec{A} \times \hat{i}\|^2 = \|\vec{A}\|^2 - (\vec{A} \cdot \hat{i})^2 = a_1^2 + a_2^2 + a_3^2 - a_1^2 = a_2^2 + a_3^2\) - \(\|\vec{A} \times \hat{j}\|^2 = \|\vec{A}\|^2 - (\vec{A} \cdot \hat{j})^2 = a_1^2 + a_2^2 + a_3^2 - a_2^2 = a_1^2 + a_3^2\) - \(\|\vec{A} \times \hat{k}\|^2 = \|\vec{A}\|^2 - (\vec{A} \cdot \hat{k})^2 = a_1^2 + a_2^2 + a_3^2 - a_3^2 = a_1^2 + a_2^2\) 5. Add these three results: $$ (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) = 2(a_1^2 + a_2^2 + a_3^2) = 2 \|\vec{A}\|^2 $$ 6. Thus the sum equals \(2 \|\vec{A}\|^2\). 7. The correct choice is (c) \(2 \|\vec{A}\|^2\).