Solution:

Let $s = AB$ and $t = AE.$ Using the $60^\circ$ angles at $A,$ $B,$ $C$ and the area formula $\frac{1}{2}xy\sin 60^\circ:$ $[AEF] = \frac{\sqrt{3}}{4} \cdot 5t,$ $[BFD] = \frac{\sqrt{3}}{4}(s-5)(s-2),$ and $[CDE] = \frac{\sqrt{3}}{4} \cdot 2(s-t).$ Subtracting all three from $[ABC] = \frac{\sqrt{3}}{4}s^2$ and simplifying, $$[DEF] = \frac{\sqrt{3}}{4}\bigl(5(s - t) + 2t - 10\bigr) = 14\sqrt{3},$$ so $5(s - t) + 2t = 66.$

At $E,$ the angles $\angle AEF$ and $\angle CED$ sum to $180^\circ - 60^\circ = 120^\circ,$ while in triangle $AEF$ the angles $\angle AEF$ and $\angle AFE$ also sum to $120^\circ.$ Hence $\angle AFE = \angle CED,$ and since $\angle A = \angle C = 60^\circ,$ triangles $AEF$ and $CDE$ are similar. Then $\frac{AE}{AF} = \frac{CD}{CE}$ gives $\frac{t}{5} = \frac{2}{s - t},$ so $t(s - t) = 10.$

Substituting $s - t = \frac{10}{t}$ into $5(s - t) + 2t = 66$ gives $\frac{50}{t} + 2t = 66,$ or $t^2 - 33t + 25 = 0,$ so $t = \frac{33 \pm \sqrt{989}}{2}.$ From $\frac{25}{t} = 33 - t$ we get $\frac{10}{t} = \frac{2}{5}(33 - t),$ so $s = t + \frac{10}{t} = \frac{3t + 66}{5} = \frac{231 \pm 3\sqrt{989}}{10}.$ Both values yield valid configurations, so $r = 989.$