Proof 1: Since $\triangle A_{1} A_{2} A_{3}$ is a non-equilateral triangle, it is easy to see that the circumcenters of $\triangle A_{1} B_{1} I$, $\triangle A_{2} B_{2} I$, and $\triangle A_{3} B_{3} I$ can all be defined.
We first look at the following lemma.
Lemma: Let the incenter of $\triangle ABC$ be $I$, and $T$ be the circumcenter of $\triangle BIC$. Then $T$ must lie on the internal angle bisector of $\angle A$.
Proof of the lemma: As shown in the figure, construct the external angle bisectors of $\angle B$ and $\angle C$, which intersect at the excenter $E$, and $E$ lies on the internal angle bisector of $\angle A$. Since $BE \perp BI$ and $CE \perp CI$, quadrilateral $BECI$ is cyclic, and its circumcenter lies on $EI$. This circumcenter is also the circumcenter of $\triangle BIC$. The lemma is proved.
We now proceed to prove the original problem.
For $i=1,2,3$, let $O_{i}$ be the center of $\odot C_{i}$, and $T_{i}$ be the circumcenter of $\triangle A_{i+1} I A_{i+2}$. Clearly, $O_{i}$ lies on the internal angle bisector of $\angle A_{i}$. By the lemma, $T_{i}$ also lies on the internal angle bisector of $\angle A_{i}$. Therefore, $\triangle O_{1} O_{2} O_{3}$ and $\triangle T_{1} T_{2} T_{3}$ are perspective from point $I$. By Desargues' theorem, they must be perspective from a line, i.e., if we denote $Q_{i} (i=1,2,3)$ as the intersection of lines $O_{i+1} O_{i+2}$ and $T_{i+1} T_{i+2}$, then $Q_{1}, Q_{2}, Q_{3}$ are collinear. Since $T_{i+1} T_{i+2}$ is the perpendicular bisector of $A_{i} I$, and $O_{i+1} O_{i+2}$ is the perpendicular bisector of $B_{i} I$, the points $Q_{1}, Q_{2}, Q_{3}$ are precisely the circumcenters of $\triangle A_{1} B_{1} I$, $\triangle A_{2} B_{2} I$, and $\triangle A_{3} B_{3} I$.
Note: Students unfamiliar with Desargues' theorem can reason as follows.
For $\triangle I O_{1} O_{2}$, $\triangle I O_{2} O_{3}$, and $\triangle I O_{3} O_{1}$, and the point sets $(T_{1}, T_{2}, Q_{3})$, $(T_{2}, T_{3}, Q_{1})$, and $(T_{3}, T_{1}, Q_{2})$, applying Menelaus' theorem, we can find some familiar results:
$$\begin{array}{l} \frac{O_{1} T_{1}}{I T_{1}} \cdot \frac{I T_{2}}{O_{2} T_{2}} \cdot \frac{O_{2} Q_{3}}{O_{1} Q_{3}}=1, \\ \frac{I T_{3}}{O_{3} T_{3}} \cdot \frac{O_{2} T_{2}}{I T_{2}} \cdot \frac{O_{3} Q_{1}}{O_{2} Q_{1}}=1, \\ \frac{I T_{1}}{O_{1} T_{1}} \cdot \frac{O_{3} T_{3}}{I T_{3}} \cdot \frac{O_{1} Q_{2}}{O_{3} Q_{2}}=1 . \end{array}$$

Multiplying the above three equations, we get
$\frac{O_{2} Q_{3}}{O_{1} Q_{3}} \cdot \frac{O_{3} Q_{1}}{O_{2} Q_{1}} \cdot \frac{O_{1} Q_{2}}{O_{3} Q_{2}}=1$.
This shows that points $Q_{1}, Q_{2}, Q_{3}$ are collinear.
Proof 2: This proof is based on inversion.
We take the incenter $I$ as the center of inversion, and the power of inversion is arbitrary. Using the symbol “,” to denote the image of a point after inversion, we get the following "dual" diagram.
In fact, the image of $\odot C_{i}$ is the line $B_{i+1}^{\prime} B_{i+2}^{\prime}$, and these lines form $\triangle B_{1}^{\prime} B_{2}^{\prime} B_{3}^{\prime}$. The line $A_{i} A_{i+1}$ is transformed into the circle $\Gamma_{i+2}$, and the side $A_{i} A_{i+1}$ is transformed into the arc not containing point $I$.
Since the distances from point $I$ to the sides of $\triangle A_{1} A_{2} A_{3}$ are equal, the radii of these circles are equal.

We note that if $\Sigma_{1}, \Sigma_{2}, \Sigma_{3}$ are three circles passing through a point $I$ and not pairwise tangent, then their centers are collinear if and only if there exists another point $J \neq I$ such that these three circles all pass through $J$.
We will apply this conclusion when $\Sigma_{i}$ is the circumcircle of $\triangle A_{i} B_{i} I$.
Since the inversion transforms $\Sigma_{i}$ into the line $A_{i}^{\prime} B_{i}^{\prime}$, the lines $A_{1}^{\prime} B_{1}^{\prime}, A_{2}^{\prime} B_{2}^{\prime}, A_{3}^{\prime} B_{3}^{\prime}$ must be concurrent. This implies that $\triangle A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime}$ and $\triangle B_{1}^{\prime} B_{2}^{\prime} B_{3}^{\prime}$ are similar, i.e., their corresponding sides are parallel. Since the radii of circles $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$ are equal, the triangle formed by their centers, $\triangle P_{1} P_{2} P_{3}$, has sides parallel to those of $\triangle B_{1}^{\prime} B_{2}^{\prime} B_{3}^{\prime}$. A similarity transformation centered at $I$ with a ratio of $\frac{1}{2}$ transforms $\triangle A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime}$ into a triangle with vertices at the midpoints of the sides of $\triangle P_{1} P_{2} P_{3}$. Therefore, the corresponding sides of $\triangle A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime}$ and $\triangle P_{1} P_{2} P_{3}$ are parallel. This leads to the conclusion.