I am quite new to graphs (and could use some practice with math in general) and couldn't find a solution with zero intersections to this problem.
Assume you have two sets of 3 nodes (set H and set L) where each node (ex. h1, h2, h3) from one set should connect to every node on the other set (so l1, l2, l3). Is there any way you could arrange the connections so that there are no intersections?
Example of one edge intersection: one edge intersection
Here is a solution: A graph $G$ is planar then $|E(G)|\leq 3|v(G)|-6$. If furthermore it is triangle free, then $|E(G)| \leq 2|V(G)|-4$. For $K_{3,3}$, It doesn't contains triangle as it doesn't contain odd cycle. So apply the second formula we get: $9\leq 2*6-4=8$ which is not possible! So it cannot be planar.
To show $|E(G)| \leq 2|V(G)|-4$, we will use Euler's equation: $|V|-|E|+|F|$=2 for planar graph. Notice that if you count the total number of edges surrounding the faces, you count each edge twice. So $\sum_{a\in F(G)}{len(a)}=2|E(G)|$. Now since the graph has no triangle, each face is bounded by at least 4 edges. So $4|F(G)|\leq\sum_{a\in F(G)}{len(a)}=2|E(G)|$. Solve for $|F|$ in Euler's equation and substitute into the previous inequality to get the desired inequality. Hope I got it right.
