Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$. Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$. So $\Delta{GBC} \sim \Delta{EAF}$.
\[[EAF] = \frac12 (AE)(EF)\sin \angle AEF  = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.\]
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}$. Therefore, the answer is $429+433+3=\boxed{865}$.
[ Equilateral](https://artofproblemsolving.com/wiki/index.php/Equilateral_triangle) $\triangle ABC$ is inscribed in a [circle](https://artofproblemsolving.com/wiki/index.php/Circle) of [radius](https://artofproblemsolving.com/wiki/index.php/Radius) $2$. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ [parallel](https://artofproblemsolving.com/wiki/index.php/Parallel) to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is [collinear](https://artofproblemsolving.com/wiki/index.php/Collinear) with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$  where $p, q,$ and $r$ are positive integers, $p$ and $r$  are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime), and $q$ is not [ divisible](https://artofproblemsolving.com/wiki/index.php/Divisibility) by the [ square](https://artofproblemsolving.com/wiki/index.php/Perfect_square) of any prime, find $p+q+r.$