jmc

Since $\angle BAD=90$ and $\angle EAF=60$, it follows that $\angle DAF+\angle BAE=90-60=30$. Rotate triangle $ADF$ $60$ degrees clockwise. Note that the image of $AF$ is $AE$. Let the image of $D$ be $D^{\prime }$. Since angles are preserved under rotation, $\angle DAF=\angle D^{\prime }AE$. It follows that $\angle D^{\prime }AE+\angle BAE=\angle D^{\prime }AB=30$. Since $\angle ADF=\angle ABE=90$, it follows that quadrilateral $ABE{D}^{\prime }$ is cyclic with circumdiameter $AE=s$ and thus circumradius $\frac{s}{2}$. Let $O$ be its circumcenter. By Inscribed Angles, $\angle BO{D}^{\prime }=2\angle BAD=60$. By the definition of circle, $OB=O{D}^{\prime }$. It follows that triangle $OB{D}^{\prime }$ is equilateral. Therefore, $B{D}^{\prime }=r=\frac{s}{2}$. Applying the Law of Cosines to triangle $AB{D}^{\prime }$, $\frac{s}{2}=\sqrt{10^2+11^2-(2)(10)(11)(\cos 30)}$. Squaring and multiplying by $\sqrt{3}$ yields $\frac{s^2\sqrt{3}}{4}=221\sqrt{3}-330⟹p+q+r=221+3+330=\boxed{554}$