If $\vec{a}, \vec{b}$ and $\vec{c}$ are three unit vectors inclined to each other at an angle $\theta$, then the maximum value of $\theta$ is

Let the angle between every pair of unit vectors be $\theta$. Then $$ \vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=\cos\theta. $$ Consider $$ |\vec{a}+\vec{b}+\vec{c}|^2 \ge 0. $$ Expanding, $$ |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2\vec{a}\cdot\vec{b}+2\vec{b}\cdot\vec{c}+2\vec{c}\cdot\vec{a} \ge 0. $$ Since the vectors are unit vectors, $$ 3+6\cos\theta \ge 0. $$ So, $$ \cos\theta \ge -\frac{1}{2}. $$ Hence, $$ \theta \le \frac{2\pi}{3}. $$ Therefore the maximum value is $\frac{2\pi}{3}$, which matches option $\text{C}$ after checking all given options.