jmc

First, observe that the radius of the circle is $12/2=6$ units. Also, $\angle AEB$ cuts off the two arcs $\widehat{AB}$ and $\widehat{CD}$, so $m\angle AEB=(m\widehat{AB}-m\widehat{CD})/2$. Subsituting $m\widehat{AB}=180^{\circ }$ and $m\angle AEB=60^{\circ }$ into this equation, we find $m\widehat{CD}=60^{\circ }$. By symmetry, $\angle AOC$ and $\angle DOB$ are congruent, so each one measures $(180-60)/2=60$ degrees. It follows that $AOC$ and $DOB$ are equilateral triangles. Therefore, we can find the area of each shaded region by subtracting the area of an equilateral triangle from the area of a sector.

The area of sector $AOC$ is $\left(\frac{m\angle AOC}{360^{\circ }}\right)\pi (\text{radius}{)}^2=\frac{1}{6}\pi (6{)}^2=6\pi$. The area of an equilateral triangle with side length $s$ is $s^2\sqrt{3}/4,$ so the area of triangle $AOC$ is $9\sqrt{3}$. In total, the area of the shaded region is $2(6\pi -9\sqrt{3})=12\pi -18\sqrt{3}.$ Therefore, $(a,b,c)=(12,18,3)$ and $a+b+c=\boxed{33}$.