AMC 12 2007b Problem 14

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 $AB$?

$\mathrm {(A)} 4 \mathrm {(B)} 3\sqrt{3} \mathrm {(C)} 6 \mathrm {(D)} 4\sqrt{3} \mathrm {(E)} 9$