Point $P$ is inside equilateral $\triangle ABC$ . Points $Q$ , $R$ , and $S$ are the feet of the perpendiculars from $P$ to $\overline{AB}$ , $\overline{BC}$ , and $\overline{CA}$ , respectively. Given that $PQ=1$ , $PR=2$ , and $PS=3$ , what is $AB$ ?