jmc

Let $D$ be the foot of the perpendicular from $O$ to the plane of $ABC$. By the Pythagorean Theorem on triangles $△OAD$, $△OBD$ and $△OCD$ we get: $$D{A}^2=D{B}^2=D{C}^2=20^2-O{D}^2$$ It follows that $DA=DB=DC$, so $D$ is the circumcenter of $△ABC$. By Heron's Formula the area of $△ABC$ is (alternatively, a $13-14-15$ triangle may be split into $9-12-15$ and $5-12-13$ right triangles): $$K=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21(21-15)(21-14)(21-13)}=84$$ From $R=\frac{abc}{4K}$, we know that the circumradius of $△ABC$ is: $$R=\frac{abc}{4K}=\frac{(13)(14)(15)}{4(84)}=\frac{65}{8}$$ Thus by the Pythagorean Theorem again, $$OD=\sqrt{20^2-R^2}=\sqrt{20^2-\frac{65^2}{8^2}}=\frac{15\sqrt{95}}{8}.$$ So the final answer is $15+95+8=\boxed{118}$.