Point $P$ is inside equilateral $\triangle ABC$. Points $Q,R,S$ are the feet of the perpendiculars from $P$ to $\overline{AB},\overline{BC},\overline{CA}$ respectively. Given that $PQ=2, PR=3,$ and $PS=5$, what is $AB$?

1. $PQ+PR+PS$ equals the height $h$ of the equilateral triangle.
2. $h=2+3+5=10$, so $AB=\dfrac{2h}{\sqrt3}=\dfrac{20}{\sqrt3}=\boxed{\dfrac{20\sqrt3}3}$.