$AB=$ the radical axis of of $S_1,S_2$ $CP=$ the radical axis of of $S_1,S_3$ $DQ=$ the radical axis of of $S_2,S_3$ The centers of $S_1,S_2,S_3$ are not collinear if $A\neq B$. So the three circles are non-coaxial. Thus their pairwise radical axises concur i.e. $AB,CP,DQ$ concur. If $A\equiv B$ then $A$ is on the line joining the centers, $C\equiv P, D\equiv Q$ and and then $AB$ is the common tangent of $S_1,S_2$, $CP$ is the tangent to $S_1$ at $C$ and $DQ$ is the tangent to $S_2$ at $D$. So $AB,CP,DQ$ are parallel, i.e. the concur at the point at infinity. Is my solution correct?

Yes your solution is correct and nicely presented. The similar thing can also be showed using power of a point but the argument is essentially the same and neater in your case.