9. As shown in the figure, three tangyuan (spherical) with a radius of $r$ are placed in a hemispherical bowl with a radius of $6 \mathrm{~cm}$. The tops of the three tangyuan are exactly coplanar with the rim of the bowl, then the radius of the tangyuan $r=$ $\qquad$ $\mathrm{cm}$.
Given the radius of the tangyuan is $r$, then
$$
\begin{array}{l}
O O_{1}=O O_{2}=O O_{3}=6-r, \\
O_{1} O_{2}=O_{1} O_{3}=O_{2} O_{3}=2 r, O M=r .
\end{array}
$$

Since $O_{1} M=\frac{2 \sqrt{3}}{3} r$, in $\triangle O O_{1} M$,
$$
\begin{array}{l}
\left(\frac{2 \sqrt{3}}{3} r\right)^{2}+r^{2}=(6-r)^{2} \\
\Rightarrow \frac{4}{3} r^{2}=36-12 r \Rightarrow r^{2}+9 r-27=0 \\
\Rightarrow r=\frac{3 \sqrt{21}-9}{2} .
\end{array}
$$