jmc

By symmetry, the areas of the two parts of the shaded region are equal. Consider the right part of the shaded region and the left triangle.



The shaded area is equal to the area of sector $PQS$ minus the area of triangle $PQS.$

Since $\angle PQS=60^{\circ }$ and $PQ=12,$ the area of sector $PQS$ is $$\frac{1}{6}\cdot 12^2\cdot \pi =24\pi .$$Also, triangle $PQS$ is equilateral with side length 12, so its area is $$\frac{\sqrt{3}}{4}\cdot 12^2=36\sqrt{3}.$$Thus, the area of the right part of the shaded region is $24\pi -36\sqrt{3},$ so the area of the entire shaded region is $$2(24\pi -36\sqrt{3})=\boxed{48\pi -72\sqrt{3}}.$$