AMC 10 A

The given situation is shown in the figure below, where $D$ is the center of the circle, $E$ is the point of tangency between the circle and the triangle, $F$ is the intersection of line $DE$ with line $l$, and $G$ is the projection of $D$ onto $l$.

Because $\angle BAC = 60^\circ$ and $\angle DEA = \angle DGF = 90^\circ$, both $\triangle AEF$ and $\triangle DGF$ are $30$-$60$-$90^\circ$ right triangles. Therefore $DF = 24$, $EF = 24 - 12 = 12$, $AE = 4\sqrt{3}$, $AF = 8\sqrt{3}$, $FG = 12\sqrt{3}$, and $AG = 12\sqrt{3} - 8\sqrt{3} = 4\sqrt{3}$. The area of kite $GAED$ is twice the area of $\triangle GAD$, so it is $48\sqrt{3}$. The area of the $60^\circ$-sector $EDG$ of the circle is $\frac{1}{6} \cdot \pi \cdot 12^2 = 24\pi$. Thus the required area, shaded in the figure, is $48\sqrt{3} - 24\pi$, and the requested sum is $48 + 3 + 24 = 75$.

OR

The required area is $\frac{1}{6}$ of the difference between the area of a circle of radius $12$ and a circumscribed regular hexagon. The hexagon is the union of $6$ equilateral triangles of side length $s = 12 \cdot \frac{2}{\sqrt{3}}$. The area of the hexagon is $6 \cdot \frac{\sqrt{3}}{4}s^2$, which equals $288\sqrt{3}$. The area of the circle is $144\pi$. Then $\frac{1}{6}$ of the difference is $48\sqrt{3} - 24\pi$, and the requested sum is $48 + 3 + 24 = 75$.