If  $\cos \alpha +\cos \beta +\cos \gamma =0=\sin \alpha +\sin \beta +\sin \gamma$  then  $\frac{\sin 3\alpha +\sin 3\beta +\sin 3\gamma }{\sin (\alpha +\beta +\gamma )}$  is equal to 

Let  $a=e^{i\alpha },c=e^{i\gamma }$  clearly  $a+b+c=0\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
 $a+b+c=0\Rightarrow a^3+b^3+c^3=3abc$
$\Rightarrow \sin 3\alpha +\sin 3\beta +\sin 3\gamma =3\sin (\alpha +\beta +\gamma )$  .