I presume $Q$ is the common point of the three circles. Cute problem. Invert about $Q$. Then, the three circles turn into lines that bound a triangle $A'B'C'$. Because $A,B,C$ are collinear, $Q$ lies on the circumcircle of $A'B'C'$. The centers of the three circles are the reflections of $Q$ across $A'B', B'C', C'A'$. The feet of the perpendiculars from $Q$ to $A'B', B'C', C'A'$ are collinear by Simson's theorem so by a homothecy centered at $Q$ with ratio $2$, the three circle centers are collinear. Thus, the centers of the three circles and $Q$ (before inverting) are concyclic, as desired. Comment. The problem can actually be strengthened to the following: $A,B,C$ are collinear if and only if the centers of the three circles and $Q$ are concyclic. This easily follows from the iff condition of Simson's theorem.