Fifteen distinct points are designated on $\triangle ABC$: the 3 vertices $A$, $B$, and $C$; $3$ other points on side $\overline{AB}$; $4$ other points on side $\overline{BC}$; and $5$ other points on side $\overline{CA}$. Find the number of triangles with positive area whose vertices are among these $15$ points.
Every triangle is uniquely determined by 3 points. There are $\binom{15}{3}=455$ ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are $\binom{5}{3}$ invalid cases on segment $AB$, $\binom{6}{3}$ invalid cases on segment $BC$, and $\binom{7}{3}$ invalid cases on segment $CA$ for a total of $65$ invalid cases. The answer is thus $455-65=\boxed{390}$.