jmc

Adding the given equations gives $2(ab+bc+ca)=484$, so $ab+bc+ca=242$. Subtracting from this each of the given equations yields $bc=90$, $ca=80$, and $ab=72$. It follows that $a^2{b}^2{c}^2=90\cdot 80\cdot 72=720^2$. Since $abc>0$, we have $abc=\boxed{720}$.