smc

Because the line $x+y=7$ is parallel to $\overline{AB}$, the area of $△ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$. In that case the triangle has base $AC=4$ and altitude $3$, so its area is $\frac{1}{2}\cdot 4\cdot 3=\boxed{6}$.