From any point $X_0$ on $\gamma_0$ draw lines $X_0A_2X_1$ and $X_0A_1X_2$, where $X_1$ lies on $\gamma_1$ and $X_2$ lies on $\gamma_2$. The points $X_1, A_0, X_2$ are collinear, and the triangle $X_0X_1X_2$ is an equilateral triangle satisfying the conditions in the statement.

Let $M_k$ be the midpoint of the minor arc $A_{k+1}A_{k+2}$ of the circle $\gamma_k$, and notice that the triangle $M_0M_1M_2$ is equilateral, since the $M_k$ are the centers of the inner Napoleon triangles associated with the triangle $A_0A_1A_2$. The center $X$ of the triangle $X_0X_1X_2$ is the intersection of $X_0M_0$ and $X_1M_1$, which must intersect at $60^\circ$. Since the locus of $X$ includes the three points $M_k$, it turns out that the locus of $X$ is the circle $M_0M_1M_2$.
Similarly, another circle is obtained by starting with the inner Napoleon triangles. The required locus is a pair of circles.