Without loss of generality, assume $a < b < c$. Comparing the figures and considering the change in orientation gives rise to the equations $3a + 1 = 2b$, $3b + 1 = 2c$, and $3c + 1 = 7a$. To solve this system of linear equations, use the first two equations to write $a$ and $c$ in terms of $b$, namely $a = \frac{2}{3}b - \frac{1}{3}$ and $c = \frac{3}{2}b + \frac{1}{2}$. Substituting these into the third equation gives $\frac{9}{2}b + \frac{3}{2} + 1 = \frac{14}{3}b - \frac{7}{3}$. Multiplying both sides by 6 yields $27b + 9 + 6 = 28b - 14$, which shows that $b = 29$. Back substituting then gives $a = 19$ and $c = 44$. The requested sum is $19 + 29 + 44 = 92$.