A first attempt to solve this problem would be to set all those fractions equal to a number $ k\in\mathbb{R}$ and solving the system of equations. One would get $ 3(5k-2)(k-2)a = (5-3k)(k-2)$. The first observation is that if $ k = 2$, this equation is always satisfied and so we might suspect that in the case $ k=2$ there are infinitely many solutions. Solving with $ k = 2$, indeed, gives the solutions $ (a,b,c) = (x,2x,1-4x)$ for all $ x\in\mathbb{R}$.