If $\vec{a}, \vec{b}$ and $\vec{c}$ are three unit vectors such that $\vec{a}+\vec{b}+\vec{c}=0$, then the value of $\vec{a} . \vec{b}+\vec{b} . \vec{c}+\vec{c} . \vec{a}$ is

$-\frac{3}{2}$

$\vec{a}+\vec{b}+\vec{c}=0$

so  $(\vec{a}+\vec{b}+\vec{c})^2 =0^2$

so  $|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} . \vec{b}+\vec{b} . \vec{c}+\vec{c} . \vec{a})=0$

$2(\vec{a} . \vec{b}+\vec{b} . \vec{c}+\vec{c} . \vec{a}) = -3$

$\vec{a} . \vec{b}+\vec{b} . \vec{c}+\vec{c} . \vec{a}=-\frac{3}{2}$

Option: 2