In $\triangle ABC$, $AB=13$, $BC=14$, $CA=15$, and $\overline{AH}$ is an altitude. Points $D$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$, respectively, so that $\overline{BD}$ and $\overline{CE}$ are angle bisectors, intersecting $\overline{AH}$ at $Q$ and $P$, respectively. What is $PQ$?

The semi-perimeter is $s=21$, and $[ABC]=\sqrt{21\cdot 8\cdot 7\cdot 6}=\sqrt{7056}=84$. So $AH=\dfrac{2[ABC]}{14}=12$.
$BH=\sqrt{13^2-12^2}=5$, $CH=14-5=9$.
By the angle bisector theorem in $\triangle ABH$, for the bisector $BQ$: $AQ:QH=AB:BH=13:5$, so $AQ=\dfrac{13}{18}AH$.
In $\triangle ACH$, for the bisector $CP$: $AP:PH=AC:CH=15:9=5:3$, so $AP=\dfrac58AH$.
$PQ=\left(\dfrac{13}{18}-\dfrac58\right)AH=\dfrac{52-45}{72}AH=\dfrac{7}{72}(12)=\boxed{\dfrac76}$.