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(D) Let the magnitude of the vectors be $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$. Since they are mutually perpendicular,$\vec{a} \cdot \vec{b} = \vec{b

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$4$

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(D) Let the magnitude of the vectors be $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$. Since they are mutually perpendicular,$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$.Let $\vec{v} = \vec{a} + \vec{b} + \vec{c}$. Then $|\vec{v}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3k^2$.Thus,$|\vec{v}| = \sqrt{3}k$.The angle $	heta$ between $\vec{a}$ and $\vec{v}$ is given by $\cos 	heta = \frac{\vec{a} \cdot \vec{v}}{|\vec{a}| |\vec{v}|} = \frac{\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c})}{k \cdot \sqrt{3}k} = \frac{|\vec{a}|^2 + 0 + 0}{\sqrt{3}k^2} = \frac{k^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}$.We need to find $36 \cos^2 2	heta$. Using the identity $\cos 2	heta = 2 \cos^2 	heta - 1$,we get $\cos 2	heta = 2(\frac{1}{\sqrt{3}})^2 - 1 = 2(\frac{1}{3}) - 1 = \frac{2}{3} - 1 = -\frac{1}{3}$.Therefore,$36 \cos^2 2	heta = 36(-\frac{1}{3})^2 = 36(\frac{1}{9}) = 4$.

Let $\vec{a}, \vec{b}, \vec{c}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$ with the vector $\vec{a}+\vec{b}+\vec{c}$. Then $36 \cos ^{2} 2 \theta$ is equal to $.....$

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