Triangle $ABC$ has $AB=13,BC=14$ and $AC=15$. Let $P$ be the point on $\overline{AC}$ such that $PC=10$. There are exactly two points $D$ and $E$ on line $BP$ such that quadrilaterals $ABCD$ and $ABCE$ are trapezoids. What is the distance $DE?$

$\textbf{(A) }\frac{42}5 \qquad$ $ \textbf{(B) }6\sqrt2 \qquad$ $ \textbf{(C) }\frac{84}5\qquad$ $ \textbf{(D) }12\sqrt2 \qquad$ $ \textbf{(E) }18$