Answer: 32

Small Hint:

The triangle inequality gives $30\sqrt{7} = DE \le DA + AE,$ so $DA \ge 20\sqrt{7}$

Big Hint:

The Law of Sines in triangle $ACD$ gives $DA = 20\sqrt{7}\sin\angle DCA \le 20\sqrt{7},$ so equality holds everywhere: $\angle DCA = 90^\circ$ and $E$ lies on ray $DA$ beyond $A$

Solution:

By the triangle inequality, $30\sqrt{7} = DE$ $\le DA + AE$ $= DA + 10\sqrt{7},$ so $DA \ge 20\sqrt{7}.$ On the other hand, in triangle $ACD$ the angle at $D$ is $\frac{\pi}{3}$ and $AC = 10\sqrt{21},$ so the Law of Sines gives $$\begin{aligned} DA &= \frac{AC \sin\angle DCA}{\sin\frac{\pi}{3}} \\ &= \frac{10\sqrt{21}}{\frac{\sqrt{3}}{2}}\sin\angle DCA \\ &= 20\sqrt{7}\sin\angle DCA \\ &\le 20\sqrt{7}. \end{aligned}$$

Both bounds force $DA = 20\sqrt{7},$ so $\angle DCA = 90^\circ,$ and equality in the triangle inequality means $E$ lies on line $AD$ with $A$ between $D$ and $E.$ From the right triangle, $DC = \sqrt{DA^2 - AC^2}$ $= \sqrt{2800 - 2100}$ $= 10\sqrt{7},$ and since $\angle CDF = 60^\circ,$ the foot satisfies $DF = DC\cos 60^\circ = 5\sqrt{7}.$

Points $F$ and $E$ are on line $AD$ on the same side of $D,$ so $EF = DE - DF$ $= 30\sqrt{7} - 5\sqrt{7}$ $= 25\sqrt{7},$ and $m + n = 25 + 7 = 32.$