From the given equations, we can observe and infer the following general proposition:
$${\sin }^{2}(\alpha -{60}^{\circ })+{\sin }^{2}\alpha +{\sin }^{2}(\alpha +{60}^{\circ })=\frac{3}{2}$$

Let's prove this proposition. The left side of the equation can be rewritten as:

$$\begin{aligned} \text{Left side} &=\frac{1-\cos (2\alpha -120^{\circ })}{2}+\frac{1-\cos 2\alpha }{2}+\frac{1-\cos (2\alpha +120^{\circ })}{2} \\ &=\frac{3}{2}-\frac{1}{2}\left[ \cos (2\alpha -120{}^\circ )+\cos 2\alpha +\cos (2\alpha +120{}^\circ ) \right] \\ &=\frac{3}{2} \quad \text{(since the sum of cosines equals 0)} \end{aligned}$$

This matches the right side of the equation. Therefore, the proposition is proven to be correct, and we can conclude that:

$$\boxed{{\sin }^{2}(\alpha -{60}^{\circ })+{\sin }^{2}\alpha +{\sin }^{2}(\alpha +{60}^{\circ })=\frac{3}{2}}$$