With the usual approach of constructing a right triangle, we can derive that $ \arccos \frac{y}{\sqrt{1 + y^2}} = \arctan \frac{1}{y} $ and $ \arcsin \frac{3}{\sqrt{10}} = \arctan 3, $ so \[\arctan x + \arctan \frac{1}{y} = \arctan 3.\]Then \[\tan \left( \arctan x + \arctan \frac{1}{y} \right) = 3,\]so from the angle addition formula, \[\frac{x + \frac{1}{y}}{1 - \frac{x}{y}} = 3.\]This becomes $ xy + 3x - 3y + 1 = 0, $ so $ (a,b,c) = \boxed{(3,-3,1)} $.