jmc

Label the points where the plane intersects the top face of the cylinder as $C$ and $D$, and the center of the cylinder as $O$, such that $C,O,$ and $A$ are collinear. Let $T$ be the center of the bottom face, and $M$ the midpoint of $\overline{AB}$. Then $OT=4$, $TM=3$ (because of the 120 degree angle), and so $OM=5$. Project $C$ and $D$ onto the bottom face to get $X$ and $Y$, respectively. Then the section $ABCD$ (whose area we need to find), is a stretching of the section $ABXY$ on the bottom face. The ratio of stretching is $\frac{OM}{TM}=\frac{5}{3}$, and we do not square this value when finding the area because it is only stretching in one direction. Using 30-60-90 triangles and circular sectors, we find that the area of the section $ABXY$ is $18\sqrt{3}+12\pi$. Thus, the area of section $ABCD$ is $20\pi +30\sqrt{3}$, and so our answer is $20+30+3=\boxed{53}$.