6. In space, there are four spheres with radii of 2, $2,3,3$, each sphere is tangent to the other three spheres. There is another smaller sphere that is externally tangent to these four spheres. What is the radius of this smaller sphere?
6. Solution: As shown in the figure, construct a tetrahedron $ABCD$ with the centers of the four spheres as vertices, where $AB=6, CD=4, AC=BC=AD=BD=5$. Let the midpoints of $AB$ and $CD$ be $F$ and $E$ respectively, and the center of the small sphere be $O$. By the symmetry of the figure, point $O$ lies on $EF$. Let the radius of the small sphere be $r$, then $OE=\sqrt{OC^2-EC^2}=\sqrt{(r+2)^2-2^2}=\sqrt{r(r+4)}, OF=\sqrt{OA^2-AF^2}=\sqrt{(r+3)^2-3^3}=\sqrt{r(r+6)}$.
$EF=\sqrt{AE^2-AF^2}=\sqrt{AD^2-DE^2-AF^2}=\sqrt{5^2-2^2-3^2}=\sqrt{12}$, Since $EF=OE+OF$, we have $\sqrt{12}=\sqrt{r(r+4)}+\sqrt{r(r+6)}$, solving this gives $r=\frac{6}{11}$, hence the radius of the small sphere is $\frac{6}{11}$.