Take the obvious graph interpretation. The answer is $\boxed{864}$, archived by a complete $4$-partite graph $K_{6,6,6,6}$. Now we proceed to prove the bound. The claim is Claim : Given any edge $uv$. The number of vertices $w$ which makes $u, v, w$ a $K_3$ is at least $\frac{\deg(u) + \deg(v) - 12}{2}$. Proof : Clearly the number of edges incident to exactly one of $u, v$ is $\deg(u) + \deg(v) - 2$. At most $10$ edges is incident to a vertex which connects to exactly one of $u, v$. Thus at least $\deg(u)+\deg(v)-12$ must connect to a vertex adjacent to both $u,v$, implying the claim. Sum up all these numbers over all ordered pair $(u,v)$ such that $uv\in E$. The number of $K_3$ is \begin{align*} \frac{1}{6}\sum_{uv\in E} \frac{\deg(u) + \deg(v) - 12}{2} &= \frac{1}{6}\left(\sum_{uv\in E} \deg(u) - \sum_{uv\in E} 6\right) \\ &= \frac{1}{6}\left(\sum_{u\in V} \deg(u)\sum_{v\in V, uv\in E} 1 - 12|E|\right) \\ &= \frac{1}{6}\left(\sum_{u\in V} \deg(u)^2 - 2592\right) \\ &\geqslant \frac{1}{6}\left[\frac{1}{|V|}\left(\sum_{u\in V}\deg(u)\right)^2 - 2592\right]\\ & = \frac{1}{6}\left(\frac{(2\cdot 216)^2}{24}-2592\right) \\ &= 864 \end{align*}as desired.