By the double angle formula, we have \begin{align*} &\sin^2\alpha + \sin^2\Big(\alpha + \frac{\pi}{3}\Big)+\sin^2\Big(\alpha - \frac{\pi}{3}\Big)\\ =\ &\frac{1-\cos 2\alpha}{2} + \frac{1-\cos(2\alpha +\frac{2\pi}{3})}{2} + \frac{1-\cos(2\alpha -\frac{2\pi}{3})}{2}\\ =\ &\frac{3}{2} - \frac{1}{2}\Big(\cos 2\alpha + \cos(2\alpha +\frac{2\pi}{3})+\cos(2\alpha -\frac{2\pi}{3})\Big)\\ =\ &\frac{3}{2} - 0\\ =\ &\boxed{\frac{3}{2}} \end{align*}

See # 3860 for the reason why $$\cos 2\alpha + \cos(2\alpha +\frac{2\pi}{3})+\cos(2\alpha -\frac{2\pi}{3})=0$$