6. A small ball is placed inside a rectangular container and is in contact with three faces that meet at a point. A point on the ball is $3 \mathrm{~cm}$, $3 \mathrm{~cm}$, and $6 \mathrm{~cm}$ away from these three faces, respectively. Then the radius of this ball is ( ).
(A) only $3 \mathrm{~cm}$
(B) only $6 \mathrm{~cm}$
(C) only $9 \mathrm{~cm}$
(D) none of the above statements is correct
6. (D).

For a small sphere with center $O$, a point $P$ on the sphere projects onto three planes $\alpha, \beta, \gamma$ intersecting at point $M$ as points $A, B, C$ respectively. The projection of point $O$ onto plane $\gamma$ is $O_{1}$, and $P Q \perp O O_{1}$ at $Q$.
Given that $P A = P B = 3 \text{ cm}, P C = 6 \text{ cm}$ (as shown in Figure 2).
From the given conditions, $O_{1}, C, M$ are collinear,
$$
C M = 3 \sqrt{2} \text{ cm}, O_{1} M =
$$
$R \sqrt{2} \text{ cm}$ (where $R$ is the radius of the sphere),
$$
\begin{aligned}
O_{1} C= & (R-3) \sqrt{2} \text{ cm}, \\
O Q & =\left|O_{1} O - O_{1} Q\right| \\
& =|R - P C| \\
& =|R - 6| \text{ cm} \\
P Q & =O_{1} C=(R-3) \sqrt{2} \text{ cm} .
\end{aligned}
$$

In the right triangle $\triangle O P Q$,
$$
\begin{array}{l}
O P^{2}=O Q^{2}+P Q^{2} \\
R^{2}=(R-6)^{2}+2(R-3)^{2},
\end{array}
$$

Solving this, we get $R=3$ or 9.