One can guess, by drawing a good diagram, that the common point is $O$ and that the lines $A_2A_3$, $B_1B_3$ and $C_1C_2$ are the perpendicular bisectors of the triangle $ABC$. So it is sufficient to prove that $A_2A_3$ is the perpendicular bisector of $BC$. For the sake of simplicity, let $\angle A = \angle BAC$, $\angle B = \angle ABC$ and $\angle C = \angle ACB$. Then, considering that the perpendicular bisector of $BC$ also bisects the angle $\angle BOC$, it suffices to prove that the angle between either the lines $OA_2$, $OA_3$ and either the lines $OB$, $OC$ is $\angle A$ or $180^\circ - \angle A$.

By the definition of $A_2$ and $A_3$, $O, A, A_2, C$ lies on the same circle, as well as $O, A, A_3, B$. Now, if $A$ and $A_2$ are opposite vertices in the quadrilateral $AOA_2C$ then $\angle A_2OC = \angle A_2AC$ and, since $A_2$ lies on $AB$, $\angle A_2AC = \angle A$ or $\angle A_2AC = 180^\circ - \angle A$. Either way, we are done. If $A$ and $A_2$ are neighbors in the quadrilateral $AA_2OC$, then $\angle A_2OC = 180^\circ - \angle A_2AC$ and we are done again. Finally, if $A$ and $A_2$ are neighbors in the quadrilateral $AA_2CO$, then $\angle A_2OC = \angle A_2AC$ and we are done. The result follows analogously to $O, A, A_3, B$, so all cases are covered.

Another solution can be found by applying an inversion with respect to the circumcircle of $ABC$. It can be shown that it maps $A_2$ to $A_3$, $B_1$ to $B_3$ and $C_1$ to $C_2$, proving directly that all lines $A_2A_3$, $B_1B_3$ and $C_1C_2$ pass through the center of inversion $O$.