jmc

Let $C$ have coordinates $(p,q)$. Then by the Shoelace Formula, the area of $△ABC$ is $\frac{3}{2}|12q-5p|$. Since $p$ and $q$ are integers, $|12q-5p|$ is a positive integer, and by Bezout's Lemma, it can equal $1$ (e.g. with $q=2,p=5$), so the minimum area is $\frac{3}{2}\times 1=\boxed{\frac{3}{2}}$.