The equation will be \[\frac{180^{\circ}(a-2)}{a} + \frac{180^{\circ}(b-2)}{b} + \frac{180^{\circ}(c-2)}{c} = 360^{\circ}.\]Dividing both sides by $180^{\circ}$, the equation becomes \[\frac{a-2}{a} + \frac{b-2}{b} + \frac{c-2}{c} = 2.\]Since $(x-2)/x = 1 - 2/x$, this further simplifies to \[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}.\]Since $a<b<c$, we have $1/a > 1/b > 1/c$ and the latter equation gives \[\frac{3}{a} > \frac{1}{2} > \frac{3}{c},\]or simply $a<6<c$. Therefore, the only possible values for $a$ are $\{3,4,5\}$. The rest is case-work: If $a=3$, then $\frac{1}{b} + \frac{1}{c} = \frac{1}{6}$. Using $b<c$, this yields \[\frac{2}{b} > \frac{1}{6} > \frac{2}{c},\]or simply $b<12<c$. Checking the values $b \in \{7,8,\dots,11\}$, we find the following solutions: \[(a,b,c) = (3,7,42) , (3,8,24), (3,9,18), (3,10,15).\] If $a=4$, then $\frac{1}{b} + \frac{1}{c} = \frac{1}{4}$. Using $b<c$, this yields \[\frac{2}{b} > \frac{1}{4} > \frac{2}{c},\]or simply $b<8<c$. Checking the values $b \in \{5,6,7\}$, we find the following solutions: \[(a,b,c) = (4,5,20) \quad \text{and} \quad (4,6,12).\] If $a=5$, then $\frac{1}{b} + \frac{1}{c} = \frac{3}{10}$. Using $b<c$, this yields \[\frac{2}{b} > \frac{3}{10} > \frac{2}{c},\]or simply $b<\frac{20}{3}<c$. Since $5=a<b<\frac{20}{3}$, the only possible value for $b$ is $6$, which does not produce an integer solution. Thus, the only solutions are \[\boxed{(a,b,c) = (3,7,42) , (3,8,24), (3,9,18), (3,10,15), (4,5,20), (4,6,12)}.\]