$\frac{\tan (\alpha +\beta -\gamma )}{\tan (\alpha -\beta +\gamma )}=\frac{\tan \gamma }{\tan \beta },$ and $\beta \ne \gamma ,$ then the value of $\sin 2\alpha +\sin 2\beta +\sin 2\gamma$ is

$\frac{\tan (\alpha +\beta -\gamma )+\tan (\alpha -\beta +\gamma )}{\tan (\alpha +\beta -\gamma )-\tan (\alpha -\beta +\gamma )}=$

$\frac{\tan \gamma +\tan \beta }{\tan \gamma -\tan \beta }$

$\Rightarrow \frac{\sin 2\alpha }{\sin 2(\beta -\gamma )}=\frac{\sin (\gamma +\beta )}{\sin (\gamma -\beta )}$

$\Rightarrow \frac{\sin 2\alpha }{-2\cos (\gamma -\beta )}=\sin (\gamma +\beta )$

$\Rightarrow \sin 2\alpha =-2\sin (\gamma +\beta )\cos (\gamma -\beta )$

$=-[\sin 2\gamma +\sin 2\beta ]$

$\sin 2\alpha +\sin 2\beta +\sin 2\gamma =0$