$\begin{aligned} & \text {We have } \alpha+\beta-\gamma=\pi \text {. } \\ & \text { Now } \sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma \quad=\sin ^2 \alpha+\sin (\beta-\gamma) \sin (\beta+\gamma) \\ & \left.=\sin ^2 \alpha+\sin (\pi-\alpha) \sin (\beta+\gamma) \quad \alpha+\beta-\gamma=\pi\right) \\ & =\sin ^2 \alpha+\sin \alpha \sin (\beta+\gamma)=\sin \alpha\{\sin \alpha+\sin (\beta+\gamma)\} \\ & =\sin \alpha\{\sin (\pi- \beta+\gamma)+\sin (\beta+\gamma)\} \\ & =\sin \alpha\{-\sin (\gamma-\beta)+\sin (\gamma+\beta)\} \\ & =\sin \alpha\{2 \sin \beta \cos \gamma\}=2 \sin \alpha \sin \beta \cos \gamma\end{aligned}$