Let us denote the intersections of $a_B$, $b_C$ and $c_A$ with $b_A$, $c_B$ and $a_C$ respectively by $A'$, $B'$ and $C'$. The triangle $AA_1C_2$ is similar to $ABC$, since they have a common angle and their sides are at a ratio $1:3$. Let $O_A$ be the center of the circumscribed circle around the triangle $AA_1C_2$. Then:
$$\angle O_A A_1 A = \frac{1}{2}(180^\circ - \angle A O_A A_1) = 90^\circ - \angle AC_2 A_1 = 90^\circ - \gamma$$
Since $a_B$ is perpendicular to $O_AA_1$ it follows that the angle between $a_B$ and $AB$ equals
$$180^\circ - 90^\circ - (90^\circ - \gamma) = \gamma$$
Analogously, the angle between $b_A$ and $AB$ equals $\gamma$, so that the triangle $A_1A_2A'$ is isosceles with base $A_1A_2$, that is, the perpendicular to $AB$ through $C'$ passes through the midpoint of $A_1A_2$, which is also the midpoint of $AB$. Thus, the perpendicular passes through the center of the circumscribed circle around the triangle $ABC$. From symmetry reasons, all the three perpendiculars pass through the center of the circumscribed circle, that is, they pass through a common point.