After answering this question, I played around with the figure a little bit. I found the following interesting results, but have been unable to prove them.
Let $ABC$ be a triangle with circumscribed circle $\Gamma$ and incenter $I$. The straight lines $AI$, $BI$, and $CI$ meet $\Gamma$ again at $A_1$, $B_1$, and $C_1$, respectively. Let $x_1$, $y_1$, and $z_1$ denote the circumcircles of the triangles $AIB_1$, $BIC_1$, and $CIA_1$, respectively, while $x_2$, $y_2$, and $z_2$ are the circumcircles of the triangles $AIC_1$, $BIA_1$, and $CIB_1$, respectively. Let $i\in\{1,2\}$. The circles $y_i$ and $z_i$ intersect again at $P_i$; the circles $z_i$ and $x_i$ intersect again at $Q_i$; and the circles $x_i$ and $y_i$ intersect again at $R_i$. Moreover, $y_1$ meets $z_2$ again at $P_3$, $z_1$ meets $x_2$ again at $Q_3$, and $x_1$ meets $y_2$ again at $R_3$.
(a) Show that the straight lines $Q_1R_2$, $R_1P_2$, and $P_1Q_2$ intersect at $I$.
(b) Prove that the circumcircle of the triangle $P_3Q_3R_3$ passes through $I$.
I think inversion about the incircle of the triangle $ABC$ may be a good approach, after all the images of $x_1$, $x_2$, $y_1$, $y_2$, $z_1$, and $z_2$ under this inversion become straight lines. Then, maybe, there are some theorems in projective geometry that can deal with the rest. However, so far, I have not discovered the relations between these six lines to produce any proof yet.
Below is some not-difficult-to-obtain information that may or may not help.
The lines $B_1C_1$, $C_1A_1$, and $A_1B_1$ are the perpendicular bisectors of $AI$, $BI$, and $CI$, respectively.
Denote by $A_2$ and $A_3$ the second intersections of $AB$ with $y_1$ and $AC$ with $z_2$, respectively; denote by $B_2$ and $B_3$ the second intersections of $BC$ with $z_1$ and $BA$ with $x_2$, respectively; denote by $C_2$ and $C_3$ the second intersections of$CA$ with $x_1$ and $CB$ with $y_2$, respectively. Then, $A_2,A_3\in B_1C_1$, $B_2B_3\in C_1A_1$, and $C_2,C_3\in A_1B_1$.
The line $B_3C_2$ passes through $I$ and is parallel to $BC$. The line $C_3A_2$ passes through $I$ and is parallel to $CA$. The line $A_3B_2$ passes through $I$ and is parallel to $AB$.
The quadrilaterals $AA_2IA_3$, $BB_2IB_3$, and $CC_2IC_3$ are rhombi.
Here is a proof of parts a) and b) using inversion about the incircle.
Let $D,E,F$ be the incircle tangency points to $BC$, $CA$, and $AB$, and given a point $X$, let its image under inversion about the incircle be $X'$ (so $A\to A'$, etc).
We have that $A'$ is the midpoint of $EF$, et cetera. We now examine $A_1'$. It must lie on circle $(A'B'C')$, the nine-point circle of $DEF$, and it also lies on line $A'I$. In particular, $$I=A'A_1'\cap B'B_1' \cap C'C_1'.$$ Also, $$x_1\to A'B_1',\ z_1\to C'A_1',\ x_2\to A'C_1',\ y_2\to B'A_1'.$$ To show part a), it suffices to show that $Q_1R_2$ passes through $I$, as the other relations follow by cyclic permutations of the vertex names. Apply Pascal's theorem on the cyclic hexagon $$(B_1'A'C_1'C'A_1'B').$$ This gives that $$B_1'A'\cap C'A_1',\ A'C_1'\cap A_1'B',\ C_1'C'\cap B'B_1'$$ are collinear; the first point is $x_1'\cap z_1'=Q_1'$, the second is $x_2'\cap y_2'=R_2'$, and the third is simply $I$, finishing the proof of part a).
Now, for part b), note that $$P_3=(BIC_1)\cap (CIB_1)\implies P_3'=B'C_1'\cap C'B_1'.$$ Now, $I$ lies on $(P_3Q_3R_3)$ if and only if $P_3'$, $Q_3'$, and $R_3'$ are collinear; this is simply the axial perspectivity of triangles $A'B'C'$ and $A_1'B_1'C_1'$. By Desargue's Theorem, this is equivalent to the central perspectivity of these two triangles, which happens if and only if $A'A_1'$, $B'B_1'$, and $C'C_1'$ concur. They do in fact; they concur at $I$.