sometimes may not,for example,if two circles are too close and the other is too distance,then the three circles can be seen as three points,you can not find a equilateral triangle.but sometimes can,you can try to find it!!!!

Pretty easy. It's not hard to prove this simple lemma: Lemma Let $A$ be a fixed vertex of $\triangle ABC$ and let vertex $B$ describe a circle such that $\triangle ABC$ is similar to another given triangle. Then vertex $C$ also describes a circle. Choose some fixed point $A$ on the first circle. Then as some point $B$ describes the second circle, vertex $C$ describes another circle. Then this circle can intersect the third given circle at at most two points. Hence there can be $0,1$ or $2$ solutions.

@Diehard: The problem asks to construct all the solutions for the three given circles; your solution constructs the solutions for the equilateral triangles with a fixed vertex, that is, A. You must take care somehow for A to go round its given circle too, rather that fix it.

How to prove Lemma Let $A$ be a fixed vertex of $\triangle ABC$ and let vertex $B$ describe a circle such that $\triangle ABC$ is similar to another given triangle. Then vertex $C$ also describes a circle.

Well, if the three circles are $C_1,C_2,C_3$, then as already pointed out, we can take any $A\in C_1$ and rotate $C_2$ by $60^\circ$ around $A$. If the point which we rotate around, $A$, changes as well, the resulting locus will be an annulus, which can be easily shown by complex numbers. Now intersect the two annuli (one for each direction of rotation) with $C_3$ to obtain all solutions.