Let A, B, C be the centers of the 3 congruent circles (A), (B), (C) intersecting at the point O. Since the point O is equidistant from the centers A, B, C, it is the circumcenter of the triangle $\triangle ABC$. Let a, b, c be the common external tangents of the circle pairs (B), (C); (C), (A); (A), (B), neither of them intersecting the remaining circle and let the lines a, b, c intersect at points $A' \equiv b \cap c,\ B' \equiv c \cap a,\ C' \equiv a \cap b$, forming a triangle $\triangle A'B'C'$. Let O' be the circumcenter of this new triangle. Since the common external tangent of 2 congruent circles is parallel to their center line, $a \equiv B'C' \parallel BC,\ b \equiv C'A' \parallel CA,\ c \equiv A'B' \parallel AB$. Thus the triangles $\triangle A'B'C' \sim \triangle ABC$ are centrally similar, having the corresponding sides parallel. The lines A'A, B'B, C'C connecting the corresponding vertices of the 2 triangles meet at their homothety center. But the lines A'A, B'B, C'C are the bisectors of the angles $\angle A', \angle B', \angle C'$ (and also of the angles $\angle A, \angle B, \angle C$), hence, the homothety center is the common incenter I of the triangles $\triangle A'B'C',\ \triangle ABC$. The circumcenters O', O are the corresponding points of these 2 centrally similar triangles, hence, the line O'O also passes through the homothety center I.

yetti wrote: Let A, B, C be the centers of the 3 congruent circles (A), (B), (C) intersecting at the point O. Since the point O is equidistant from the centers A, B, C, it is the circumcenter of the triangle $\triangle ABC$. Let a, b, c be the common external tangents of the circle pairs (B), (C); (C), (A); (A), (B), neither of them intersecting the remaining circle and let the lines a, b, c intersect at points $A' \equiv b \cap c,\ B' \equiv c \cap a,\ C' \equiv a \cap b$, forming a triangle $\triangle A'B'C'$. Let O' be the circumcenter of this new triangle. Since the common external tangent of 2 congruent circles is parallel to their center line, $a \equiv B'C' \parallel BC,\ b \equiv C'A' \parallel CA,\ c \equiv A'B' \parallel AB$. Thus the triangles $\triangle A'B'C' \sim \triangle ABC$ are centrally similar, having the corresponding sides parallel. The lines A'A, B'B, C'C connecting the corresponding vertices of the 2 triangles meet at their homothety center. But the lines A'A, B'B, C'C are the bisectors of the angles $\angle A', \angle B', \angle C'$ (and also of the angles $\angle A, \angle B, \angle C$), hence, the homothety center is the common incenter I of the triangles $\triangle A'B'C',\ \triangle ABC$. The circumcenters O', O are the corresponding points of these 2 centrally similar triangles, hence, the line O'O also passes through the homothety center I. Why do A'A, B'B, C'C meet at the incentre?

Because A', B', C' are equidistant from the sides so they all lie on the respective angle bisectors.

Found an instant solve with homothety so just putting it here let $A, B, C$ be the vertices of the given triangle, and let $O_1, O_2, O_3$ be the centers of the three circles, such that the two edges passing $A$ are tangent to $(O_1)$, two edges passing $B$ are tangent to $(O_2)$, and similarly for $C$ and $O_3$. Since $O_1O_2 \parallel AB$, $O_2O_3 \parallel BC$, $O_1O_3 \parallel AC$, we get that there exists a homothety $\omega$ that maps $\triangle O_1O_2O_3$ to $\triangle ABC$. We also know that $O$ is the circumcenter of $\triangle O_1O_2O_3$, so $\omega$ takes $O$ to the circumcenter of $\triangle ABC$. Finally, notice that the two triangles $\triangle O_1O_2O_3$ and $\triangle ABC$ have the same incenter $I$, so $I$ is the centre of homothety which is collinear with the two circumcenters.