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(B) Given that $\vec{a} \cdot \vec{b} = 0$,$\vec{b} \cdot \vec{c} = 0$,$|\vec{a}| = 2$,$|\vec{b}| = 3$,$|\vec{c}| = 5$ and $|\vec{a} + \vec{b} + \vec{

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$\frac{\pi}{3}$

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(B) Given that $\vec{a} \cdot \vec{b} = 0$,$\vec{b} \cdot \vec{c} = 0$,$|\vec{a}| = 2$,$|\vec{b}| = 3$,$|\vec{c}| = 5$ and $|\vec{a} + \vec{b} + \vec{c}| = 4 \sqrt{3}$.We know that $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.Substituting the given values:$(4 \sqrt{3})^2 = 2^2 + 3^2 + 5^2 + 2(0 + 0 + |\vec{c}| |\vec{a}| \cos 	heta)$.$48 = 4 + 9 + 25 + 2(5)(2) \cos 	heta$.$48 = 38 + 20 \cos 	heta$.$10 = 20 \cos 	heta$.$\cos 	heta = \frac{10}{20} = \frac{1}{2}$.Therefore,$	heta = \cos^{-1} \left(\frac{1}{2}ight) = \frac{\pi}{3}$.

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{b}$ is perpendicular to $\vec{c}$. If $|\vec{a}|=2, |\vec{b}|=3, |\vec{c}|=5$ and $|\vec{a}+\vec{b}+\vec{c}|=4 \sqrt{3}$,then the angle between $\vec{a}$ and $\vec{c}$ is

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