Step 1:
Consider $(\cos^2\alpha-\sin^2\alpha)+(\cos^2\beta-\sin^2\beta)+(\cos^2\gamma-\sin^2\gamma)=\cos 2\alpha+\cos 2\beta+\cos 2\gamma=0$
We know that $\cos 2\alpha+\cos 2\beta+\cos 2\gamma=0$
Step 2:
Now $\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\sin^2\alpha+\sin^2\beta+\sin^2\gamma=(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)+(\cos^2\gamma+\sin^2\gamma)=3$
Therefore $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=\sin^2\alpha+\sin^2\beta+\sin^2\gamma=\large\frac{3}{2}$
Hence proved.