Solution:

A sphere of radius $r$ tangent to both planes has its center on the half-plane bisecting the dihedral angle. If the dihedral angle is $2\theta,$ the center is at distance $\frac{r}{\sin\theta}$ from $\ell.$ In the cross-section through the center perpendicular to $\ell,$ the point of $\ell,$ the center, and the tangent point on $\mathcal{P}$ form a right triangle with angle $\theta$ at $\ell,$ so the tangent point lies at distance $\frac{r\cos\theta}{\sin\theta}$ from $\ell.$

Measure positions along $\ell.$ The centers of the two radius-$36$ spheres are both at distance $\frac{36}{\sin\theta}$ from $\ell$ and are $72$ apart, so they differ by $72$ along $\ell,$ and by symmetry the radius-$13$ center sits halfway between them along $\ell,$ at distance $\frac{13}{\sin\theta}$ from $\ell.$ External tangency makes its distance to each big center $49:$ $$\left(\frac{36}{\sin\theta} - \frac{13}{\sin\theta}\right)^2 + 36^2 = 49^2,$$ so $\left(\frac{23}{\sin\theta}\right)^2 = 49^2 - 36^2 = 1105.$ Since $23^2 + 24^2 = 1105,$ we get $\sin\theta = \frac{23}{\sqrt{1105}}$ and $\cos\theta = \frac{24}{\sqrt{1105}}.$

The required distance is $\frac{13\cos\theta}{\sin\theta} = \frac{13 \cdot 24}{23} = \frac{312}{23},$ which is in lowest terms, so $m + n = 312 + 23 = 335.$