imc

Let $P,Q,R,X,Y,$ and $Z$ be the intersections $\overleftrightarrow{AB}\cap \overleftrightarrow{CD},\overleftrightarrow{CD}\cap \overleftrightarrow{EF},\overleftrightarrow{EF}\cap \overleftrightarrow{AB},\overleftrightarrow{BC}\cap \overleftrightarrow{DE},\overleftrightarrow{DE}\cap \overleftrightarrow{FA},$ and $\overleftrightarrow{FA}\cap \overleftrightarrow{BC},$ respectively. The sum of the interior angles of any hexagon is $720^{\circ }.$ Since hexagon $ABCDEF$ is equiangular, each of its interior angles is $720^{\circ }÷6=120^{\circ }.$ By angle chasing, we conclude that the interior angles of $△PBC,△QDE,△RFA,△XCD,△YEF,$ and $△ZAB$ are all $60^{\circ }.$ Therefore, these triangles are all equilateral triangles, from which $△PQR$ and $△XYZ$ are both equilateral triangles. We are given that \begin{alignat}{8} [PQR]&=\frac{\sqrt{3}}{4}\cdot PQ^2&&=192\sqrt3, \\ [XYZ]&=\frac{\sqrt{3}}{4}\cdot YZ^2&&=324\sqrt3, \end{alignat} so we get $PQ=16\sqrt{3}$ and $YZ=36,$ respectively. By equilateral triangles and segment addition, we find the perimeter of hexagon $ABCDEF:$ $$\begin{array}{rl}AB+BC+CD+DE+EF+FA& =AZ+PC+CD+DQ+YF+FA\\ & =(YF+FA+AZ)+(PC+CD+DQ)\\ & =YZ+PQ\\ & =36+16\sqrt{3}.\end{array}$$ Finally, the answer is $36+16+3=\boxed{55}.$