$ABC$ is a right triangle with $m\angle C=90^{\circ}$. $AC=10\text{ units}^2$ and $BC=24\text{ units}^2$. Point $P$ is located inside $ABC$ such that the distance from $P$ to $AB$ is twice the distance from $P$ to $AC$, and the distance from $P$ to $AC$ is twice the distance from $P$ to $BC$. In $\text{units}^2$, what is the distance from $P$ to $AB$?
