Let $AB = c$, $AC = b$, WLOG $BC = a$ is the longest side, and WLOG set $D$ such that $2BD = CD$. If the line passed through A then $[ABD] = 2[ACD]$, so the line must intersect line segment $AB$. Let the intersection be at $X$; then we have \[\frac{1}{2} \cdot BX \cdot \frac{2}{3}a \cdot \sin{(\angle ABC)} = \frac{1}{2} \cdot \frac{1}{2}ac\sin{(\angle ABC)}\] It follows that $BX = \frac{3}{4}c$. Then \[\frac{3c}{4} + \frac{2a}{3} = \frac{a}{3}+b+\frac{c}{4} \Longrightarrow 3c + 2a = 6b\] Taking mod 3 we see $a \equiv 0 \pmod{3}$. Testing a few values, we see the smallest possible value of $a$ is $9$, with $c=8$ and $b=7$. The next smallest possible value of $a$ is $12$, with $c=6$ and $b=7$. In both cases $abc=504$. For larger values of $a$, $abc$ is larger than 504 because of restraints caused by the Triangle Inequality. This feels too easy, but I'm not sure where I'm wrong...