Inside a right circular cone with base radius $2\sqrt3$ and height $6$ are three congruent spheres of radius $r$. Each sphere is tangent to the other two spheres and also tangent to the base and to the lateral surface of the cone. What is $r$?

1. The three centers are at height $r$ above the base, at the vertices of an equilateral triangle of side $2r$ centered on the axis. So each center is at distance $d=\frac{2r}{\sqrt3}$ from the axis.
2. In the plane through the axis and one center, the side of the cone is the line $\frac{x}{2\sqrt3}+\frac z6=1$. Its distance from the point $(d,r)$ must be $r$. Since $\sqrt{\frac1{12}+\frac1{36}}=\frac13$, this says $3\left(1-\frac{d}{2\sqrt3}-\frac r6\right)=r$.
3. Here $\frac d{2\sqrt3}=\frac r3$, so $3\left(1-\frac r3-\frac r6\right)=3-\frac{3r}2=r$.
4. Therefore $r=\boxed{\dfrac65}$.