(4) Let $A, B, C, D$ be four points on a sphere with center $O$, such that $AB, AC$, and $AD$ are mutually perpendicular, and $AB=3 \text{ cm}, AC=4 \text{ cm}, AD=\sqrt{11} \text{ cm}$. Then the radius of the sphere is ( ).
(A) $3 \text{ cm}$
(B) $4 \text{ cm}$
(C) $5 \text{ cm}$
(D) $6 \text{ cm}$
(4) Method 1 Establish a spatial rectangular coordinate system with point $A$ as the origin, and the rays $A B, A C, A D$ as the $x$-axis, $y$-axis, and $z$-axis, respectively. Then the coordinates of $B, C, D$ are $B(3,0,0), C(0,4,0), D(0,0, \sqrt{11})$. Clearly, the center of the sphere should be the intersection of the following three planes:
$$
\alpha: x=\frac{3}{2} ; \quad \beta: y=2 ; \quad \gamma: z=\frac{\sqrt{11}}{2}
$$

Thus, the coordinates of the center of the sphere are $O\left(\frac{3}{2}, 2, \frac{\sqrt{11}}{2}\right)$, and the radius of the sphere $r=O A=$ 3(cm).

Method 2 Establish a spatial rectangular coordinate system with point $A$ as the origin, and the rays $A B, A C, A D$ as the $x$-axis, $y$-axis, and $z$-axis, respectively. Then the coordinates of $B, C, D$ are $B(3,0,0), C(0,4,0), D(0,0, \sqrt{11})$. Let the equation of the sphere be
$$
\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}+\left(z-z_{0}\right)^{2}=r^{2},
$$

where $O\left(x_{0}, y_{0}, z_{0}\right)$ is the center of the sphere, and $r$ is the radius of the sphere. Since $B, C, D$ are all on the sphere, we get the system of equations
$$
\left\{\begin{array}{l}
x_{0}^{2}+y_{0}^{2}+z_{0}^{2}=r^{2}, \\
\left(x_{0}-3\right)^{2}+y_{0}^{2}+z_{0}^{2}=r^{2}, \\
x_{0}^{2}+\left(y_{0}-4\right)^{2}+z_{0}^{2}=r^{2}, \\
x_{0}^{2}+y_{0}^{2}+\left(z_{0}-\sqrt{11}\right)^{2}=r^{2} .
\end{array}\right.
$$

Solving these equations, we get $x_{0}=\frac{3}{2}, y_{0}=2, z_{0}=\frac{\sqrt{11}}{2}, r=3(\mathrm{~cm})$.