Triangle $ABC$ has side lengths $AC=3$, $BC=4$, and $AB=5$. Let $R$ be a point on the incircle $\omega$ of $\triangle ABC$. The altitude from $C$ to $\overline{AB}$ intersects $\omega$ at points $P$ and $Q$. Then, the greatest possible area of $\triangle PQR$ is $\frac{m\sqrt{n}}{p}$, where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p$.

Let's think step by step and output the final answer within \boxed{}.
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