Solution 1. Fix an index $i \in \{1, 2, 3\}$. Let $D_1, D_2, D_3$ be the points of tangency of the incircle of triangle $\triangle A_1A_2A_3$ with its sides $A_2A_3$, $A_3A_1$, $A_1A_2$ respectively.
The key observation is that given a line $\ell$ in the plane, the image of $\ell$ under the mapping $\tau_{i+2}(\tau_i(\tau_{i+1}(\ell)))$ is a line parallel to $\ell$. Indeed, $\ell$ is rotated thrice by angles equal to the angles of $\triangle A_1A_2A_3$, and the composition of these rotations induces a half-turn and translation on $\ell$ as the angles of $\triangle A_1A_2A_3$ add to $180^\circ$. Since $D_i$ is a fixed point of this transformation (by the chain of maps $D_i \xrightarrow{\tau_{i+1}} D_{i+2} \xrightarrow{\tau_i} D_{i+1} \xrightarrow{\tau_{i+2}} D_i$), we conclude that the line $\overline{PD_i}$ maps to the line $\overline{P_iD_i}$. But the two lines are parallel and both of them pass through $D_i$ hence they must coincide, so $D_i$ lies on $\overline{PP_i}$. Further, each rotation preserves distances, hence $P_i$ is the reflection of $P$ in $D_i$. In other words, the triangle $P_1P_2P_3$ is obtained by applying a homothety with ratio 2 and center $P$ to the triangle $D_1D_2D_3$. Thus, the radius of the circumcircle of $\triangle P_1P_2P_3$ is twice the radius of the circumcircle of $\triangle D_1D_2D_3$, i.e., twice the radius of the incircle of $\triangle A_1A_2A_3$, which is known to be at most the radius of the circumcircle $\Gamma$.

Now, for any complex number $z$, the rotation at $z_0$ with angle $\theta$ counterclockwise sends $z$ to $(z - z_0)e^{i\theta} + z_0$.
Therefore, one computes that
$$\begin{align*} \tau_{321}(z) &= \tau_3(\tau_1(\tau_2(z))) &= \tau_3(\tau_1((z-b)e^{iB} + b)) \\ &= \tau_3(ze^{i(A+B)} + be^{iA}(1 - e^{iB}) + a(1 - e^{iA})) \\ &= -z + b + c + be^{i(A+C)} + ae^{iC} - ae^{i(A+C)} - ce^{iC} \end{align*}$$
Thus, $\tau_{312}(z)+z$ is independent of $z$. Similarly $\tau_{123}(z)+z$ and $\tau_{231}(z)+z$ are also independent of $z$. Note that adding $z$ is the same as translation by $z$, hence we have shown that the circumradius of $\triangle P_1P_2P_3$ is independent of $P$.
Thus, it suffices to prove the result for $z = z_0 = a + b + c$. Let $U = -\tau_{312}(z_0)$, $V = -\tau_{123}(z_0)$, $W = -\tau_{231}(z_0)$. So, it is enough to prove that the circumradius of $\triangle UVW$ at most the radius of $\Gamma$.

Name the vertices $A_1, A_2, A_3$ as $A, B, C$ for convenience. Let the parallel line to $BC$ passing through $A$ intersect $\Gamma$ again at $K$. Similarly, define $L$ as the second intersection of the line through $B$ parallel to $CA$ and finally $M$ for $C$ parallel to $AB$.
We claim that $U$ lies on the line segment $\overrightarrow{AK}$: We have $U = a - (b-a)e^{i(A+C)} + (c-a)e^{iC}$, hence $\overrightarrow{AU}$ is parallel to $\overrightarrow{AK}$ hence $U$ lies on the line $AK$. If $AB = AC$ then $U = A$, and the claim is proven. Else suppose that $AB < AC$. Then $\overrightarrow{AU}$ points towards $K$ and $|AU| = AC - AB$, so it suffices

to show that $AK > AC - AB$. But this is clear because $KCBA$ is an isosceles trapezium, so $AB = KC$, and then triangle inequality on $\triangle KAC$ to get $KA + KC > AC$. Thus, $U \in \overline{AK}$, and similarly $V \in \overline{BL}$, $W \in \overline{CM}$. We claim that for any $U,V,W$ on the segments $AK, BL, CM$ respectively, the circumradius of $\triangle UVW$ is less than or equal to the radius of $\Gamma$. Now let $X,Y$ be two fixed points on the same side of a line $\ell$. Fix a side of $\overleftrightarrow{XY}$, and let $Z$ be a variable point on $\ell$ which always remains on this fixed side of $\overleftrightarrow{XY}$. Then the circumradius of $\triangle XYZ$ is minimized at the unique point $Z_0$ (on this fixed side of $\overleftrightarrow{XY}$) for which the circumcircle of $\triangle XYZ_0$ is tangent to $\ell$ and it is an increasing function as one goes further away from this unique point $Z_0$. Thus, the maximum circumradius of $\triangle UVW$ is achieved only if $U \in \{A,K\}$, $V \in \{B,L\}$, $W \in \{C,M\}$. For each of these, the circumradius is the radius of $\Gamma$, hence we are done. $\square$

Toss the figure on the complex plane, and let $A_1 = a$, $A_2 = b$, $A_3 = c$ without loss of generality. Let the angles of the triangle at $A_1, A_2, A_3$ be denoted by $A, B, C$.
Now, for any complex number $z$, the rotation at $z_0$ with angle $\theta$ counterclockwise sends $z$ to $(z - z_0)e^{i\theta} + z_0$.

The key observation is that given a line $\ell$ in the plane, the image of $\ell$ under the mapping $\tau_{i+2}(\tau_i(\tau_{i+1}(\ell)))$ is a line parallel to $\ell$. Indeed, $\ell$ is rotated thrice by angles equal to the angles of $\triangle A_1A_2A_3$, and the composition of these rotations induces a half-turn and translation on $\ell$ as the angles of $\triangle A_1A_2A_3$ add to $180^\circ$. Since $D_i$ is a fixed point of this transformation (by the chain of maps $D_i \xrightarrow{\tau_{i+1}} D_{i+2} \xrightarrow{\tau_i} D_{i+1} \xrightarrow{\tau_{i+2}} D_i$), we conclude that the line $\overline{PD_i}$ maps to the line $\overline{P_iD_i}$. But the two lines are parallel and both of them pass through $D_i$ hence they must coincide, so $D_i$ lies on $\overline{PP_i}$. Further, each rotation preserves distances, hence $P_i$ is the reflection of $P$ in $D_i$. In other words, the triangle $P_1P_2P_3$ is obtained by applying a homothety with ratio 2 and center $P$ to the triangle $D_1D_2D_3$. Thus, the radius of the circumcircle of $\triangle P_1P_2P_3$ is twice the radius of the circumcircle of $\triangle D_1D_2D_3$, i.e., twice the radius of the incircle of $\triangle A_1A_2A_3$, which is known to be at most the radius of the circumcircle $\Gamma$.