The foot of the perpendicular from $O$ to the plane of $ABC$ is equidistant from $A,$ $B,$ and $C$ (the slant segments to the vertices all have length $20$), so it is the circumcenter of triangle $ABC.$ Since $15^2\lt13^2+14^2,$ the triangle is acute and its circumcenter lies inside it, so this perpendicular length is also the distance from $O$ to the triangle itself.

By Heron's formula with $s = 21,$ the area is $K = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84,$ so the circumradius is $R = \frac{abc}{4K} = \frac{13 \cdot 14 \cdot 15}{336} = \frac{65}{8}.$ The distance from $O$ to the plane is $$\begin{aligned} \small \sqrt{20^2 - \left(\tfrac{65}{8}\right)^2} \\ &= \sqrt{\frac{25600 - 4225}{64}} \\ &= \frac{\sqrt{21375}}{8} \\ &= \frac{15\sqrt{95}}{8}. \end{aligned}$$

Here $\gcd(15, 8) = 1$ and $95 = 5 \cdot 19$ is squarefree, so $m + n + k = 15 + 95 + 8 = 118.$