I'd start by constructing a unit hexagon such that the vertices are placed at \(\displaystyle \sin\left(\frac{k\pi}{3}\right), \cos\left(\frac{k\pi}{3}\right)\) for \(k = 0, 1, \cdots, 5\). By doing this, we can see that \(A = (0,1)\) and \(K\) being the midpoint of \(\overline{EF}\) means that \(\displaystyle K = \left( \frac{\sqrt{3}}{2}, 0 \right)\). Then we can use the formula for the area of a triangle given three vertices:

\(\displaystyle \text{Area} \: = \: \biggr| \frac{A.x\left(K.y-L.y\right)+K.x\left(L.y-A.y\right)+L.x\left(A.y-K.y\right)}{2} \biggl|\)

The area of a unit hexagon is \(\displaystyle \frac{3\sqrt{3}}{2}\), so that means we want \(\displaystyle \triangle AKL\) to have an area of \(\displaystyle \frac{3\sqrt{3}}{5}\). Of the three options given in the problem text, deciding that \(L\) should lie on \(\overline{BC}\) is the easiest, since that fixes the x-coordinate in place, leaving us just one variable to solve for. Plugging in all of the known information and the desired area, we get an equation to solve:

\(\displaystyle \frac{3\sqrt{3}}{5} = \: \biggr| \frac{-\frac{\sqrt{3}}{2}\left(L.y-1\right)+\frac{\sqrt{3}}{2}}{2} \biggl|\)

...although now that I've written this all down, I'm not actually sure it solves the problem at hand, because it makes quite a few (unjustified) assumptions about the placement of the vertices and the area of the hexagon. Presumably, they want you to arrive at some revelation regarding how \(L\) might be positioned relative to \(B\) and \(C\), but I'm not quite seeing how just yet. In any case, I'll leave all of this as it is, in the hopes that my partial workings at least get you thinking in the right direction and then maybe you can take it the rest of the way. Good luck!