Let the side lengths, in counterclockwise order, be $a, b, c, d, e, f$. Place the hexagon on the coordinate plane with edge $a$ parallel to the $x$-axis and the intersection between edge $a$ and edge $f$ at the origin (oriented so that edge $b$ lies in the first quadrant). If you travel along all six sides of the hexagon starting from the origin, we get that the final $x$ coordinate must be $a+b / 2-c / 2-d-e / 2+f / 2=0$ by vector addition. Identical arguments tell us that we must also have $b+c / 2-d / 2-e-f / 2+a / 2=0$ and $c+d / 2-e / 2-f-a / 2+b / 2=0$. Combining these linear equations tells us that $a-d=e-b=c-f$. This is a necessary and sufficient condition for the side lengths to form an equiangular hexagon. WLOG say that $a=1$ and $b<f$ (otherwise, you can rotate/reflect it to get it to this case). Thus, we must either have $(a, b, c, d, e, f)=(1,5,3,4,2,6)$ or $(1,4,5,2,3,6)$. Calculating the areas of these two cases gets either $67 \sqrt{3} / 4$ or $65 \sqrt{3} / 4$, for a sum of $33 \sqrt{3}$.