Solution:

Here $AB = 84 + 168 = 252,$ so the semicircle has center $O$ (the midpoint of $\overline{AB}$) and radius $126.$ Since $AN = AO = ON = 126,$ triangle $AON$ is equilateral, so $\angle AON = 60^\circ$ and sector $AON$ is exactly one third of the semicircle. Likewise, if $Q$ is the foot of the perpendicular from $U$ to $\overline{DC},$ then $AU : UB = 1 : 2$ makes rectangle $AUQD$ one third of rectangle $ABCD.$

The part of $\mathcal{R}$ on the $A$-side of $\ell$ equals (sector $AON$) $-$ $[NUO]$ $+$ (rectangle $AUQD$) $+$ $[UQT].$ Since this must be one third of $\mathcal{R},$ we need $[NUO] = [UQT].$ Let $P$ be the foot of the perpendicular from $N$ to $\overline{AB}.$ In the $30$-$60$-$90$ triangle $NOP,$ $OP = 63$ and $NP = 63\sqrt{3},$ so $UP = 84 - 63 = 21$ and $UO = 126 - 84 = 42.$ Triangles $NUP$ and $TUQ$ are similar right triangles (vertical angles at $U$), so $$\frac{[UQT]}{[NUP]} = \left(\frac{UQ}{NP}\right)^2, \qquad \frac{[NUO]}{[NUP]} = \frac{UO}{UP} = 2,$$ the latter because both triangles have height $NP$ over bases $UO$ and $UP.$

Setting the two triangle areas equal gives $\left(\frac{UQ}{NP}\right)^2 = 2,$ so $$DA = UQ = NP\sqrt{2} = 63\sqrt{3} \cdot \sqrt{2} = 63\sqrt{6},$$ and $m + n = 63 + 6 = 69.$