Let $BC = 6$, $BX = 3$, $CX = 5$, and let $F$ be the midpoint of $\overline{BC}$. Let $\overline{AX} \perp \overline{BC}$ and $AF = \sqrt{247}$. If $AC$ is of the form $\sqrt{b}$ and $AB$ is of the form $\sqrt{c}$ where $b$ and $c$ are nonnegative integers, find $2c + 3b$.

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