Step 1:
$\cos\alpha+\cos\beta+\cos\gamma=0$
$\sin\alpha+\sin\beta+\sin\gamma=0$
Therefore $(\cos\alpha+i\sin\alpha)+(\cos\beta+i\sin\beta)+(\cos\gamma+i\sin\gamma)=0$
This is of the form $a+b+c=0$
Therefore $a^3+b^3+c^3=3abc$
$(\cos\alpha+i\sin\alpha)^3+(\cos\beta+i\sin\beta)^3+(\cos\gamma+i\sin\gamma)^3=3(\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta)(\cos\gamma+i\sin\gamma)$
Step 2:
LHS:
$\cos 3\alpha+i\sin 3\alpha+\cos 3\beta+i\sin 3\beta+\cos 3\gamma+i\sin 3\gamma=(\cos 3\alpha+\cos 3\beta+\cos 3\gamma)+i(\sin 3\alpha+\sin 3\beta+\sin 3\gamma)$
RHS:
$3e^{i\alpha}e^{i\beta}e^{i\gamma}=3e^{\large i(\alpha+\beta+\gamma)}$
$\Rightarrow 3[\cos(\alpha+\beta+\gamma)+i\sin(\alpha+\beta+\gamma)]$
Step 3:
Equating the imaginary parts we get
$\Rightarrow \sin 3\alpha+\sin 3\beta+\sin 3\gamma=3\sin(\alpha+\beta+\gamma)$