The Classic 13-14-15 Triangle, Sliced

In triangle $ABC$, $AB = 13$, $BC = 14$, and $CA = 15$. A line parallel to $BC$ intersects $AB$ at $P$ and $AC$ at $Q$ such that the area of trapezoid $BPQC$ is twice the area of triangle $APQ$. The length of $PQ$ can be expressed as $\dfrac{a\sqrt{b}}{c}$, where $a$, $b$, $c$ are positive integers with $\gcd(a,c) = 1$ and $b$ square-free. Find $a + b + c$.