A point P in the interior of a regular hexagon is at distance...

Note that $\mathrm{CF}=2 \mathrm{AB}, \mathrm{PA}=2 \mathrm{PC}$ \& $\mathrm{PB}=2 \mathrm{PF}$ Hence $\triangle \mathrm{PAB}$ is similar to $\triangle \mathrm{PFC}$, hence $\mathrm{A}_1 \mathrm{P}_1 \mathrm{C}$ \& $\mathrm{B}_1 \mathrm{P}_1 \mathrm{~F}$ are collinear. Let each side of hexagon be equal to $x$. Let, $Q \& R$ be foot of altitudes from $P$ to base $A B \& C F$ respectively. So $R$ is centre of hexagon Now $\frac{1}{3} \times \frac{\sqrt{3} x}{2}=\sqrt{64-\frac{x^2}{4}}$ $\Rightarrow \frac{x^2}{12}=64-\frac{x^2}{4}$ $\Rightarrow \frac{4 x^2}{12}=64 \Rightarrow x=8 \sqrt{3}$ Note that circumradius of a regular hexagon = side of regular hexagon Hence $r=8 \sqrt{3} \approx 13.856$ Hence nearest integer $=14$