If $\sin \alpha +\sin \beta +\sin \gamma =0$ and $\cos \alpha +\cos \beta +\cos \gamma =0$, then value of $\cos (\alpha -\beta )+\cos (\beta -\gamma )+\cos (\gamma +\alpha )$ is

$\sin \alpha +\sin \beta +\sin \gamma =0\Rightarrow \sin \alpha +\sin \beta =-\sin \gamma$   ...(1)
$\cos \alpha +\cos \beta +\cos \gamma =0\Rightarrow \cos \alpha +\cos \beta =-\cos \gamma$   ...(2)
Squaring and adding (1) and (2), we get
$$\begin{gathered} {\left(\sin \alpha +\sin \beta \right)}^2+{\left(\cos \alpha +\cos \beta \right)}^2={\sin }^2\gamma +{\cos }^2\gamma \\ \Rightarrow 2+2\cos \left(\alpha -\beta \right)=1\Rightarrow \cos \left(\alpha -\beta \right)=-\frac{1}{2} \end{gathered}$$
Similarly, $\cos \left(\beta -\gamma \right)=-\frac{1}{2},\cos \left(\gamma -\alpha \right)=-\frac{1}{2}$
Therefore,
$\cos \left(\alpha -\beta \right)+\cos \left(\beta -\gamma \right)+\cos \left(\gamma -\alpha \right)=-\frac{3}{2}$