3. In space, there are 4 spheres with radii of $2,2,3,3$, each sphere is externally tangent to the other 3 spheres. Additionally, there is a smaller sphere that is externally tangent to these 4 spheres. Find the radius of the smaller sphere.
(CMO - 10 Problem)
3. Let the centers of the four spheres be $A, B, C, D$. According to the problem, we have $A B=6, C D=4, A C=B C=A D=B D=5$.

Let the midpoints of $A B$ and $C D$ be $F$ and $E$, respectively, and let the center of the small sphere be $O$. By symmetry, $O$ lies on the line segment $E F$, and it is easy to see that $E F \perp A B, E F \perp C D$. Therefore, $E F=\sqrt{A E^{2}-A F^{2}}=\sqrt{A C^{2}-C E^{2}-A F^{2}}=\sqrt{12}, O E=\sqrt{O \bar{C}^{2}-C E^{2}}=$ $\sqrt{r^{2}+4 r}, O F=\sqrt{O A^{2}-A F^{2}}=\sqrt{r^{2}+6 r}$, where $r$ is the radius of the small sphere $O$. Thus, we have $\sqrt{r^{2}+4 r}+\sqrt{r^{2}+6 r}=$ $\sqrt{12}$, which simplifies to $\sqrt{r+4}+\sqrt{r+6}=\sqrt{\frac{12}{r}}$. Also, $(\sqrt{r+6})^{2}-(\sqrt{r+4})^{2}=2$, so $\sqrt{r+6}-\sqrt{r+4}=\sqrt{\frac{r}{3}}$. Therefore, $2 \sqrt{r+6}=\frac{1}{\sqrt{3 r}}(r+6)$. Since $r>0$, we have $2 \sqrt{3} r=\sqrt{r+6}$, which simplifies to $12 r=r+6$, hence $r=\frac{6}{11}$ is the solution.