$23 \cdot 22$ Among the following groups of numbers, one group cannot be the four heights of a certain tetrahedron. This group is
(A) $1, \frac{3 \sqrt{2}}{2}, \frac{3 \sqrt{2}}{2}, \frac{3 \sqrt{2}}{2}$.
(B) $4, \frac{25 \sqrt{3}}{3}, \frac{25 \sqrt{3}}{3}, \frac{25 \sqrt{3}}{3}$.
(C) $\sqrt{5}, \frac{\sqrt{30}}{2}, \frac{\sqrt{30}}{2}, \frac{\sqrt{30}}{2}$.
(D) $2, \frac{6 \sqrt{5}}{5}, \frac{6 \sqrt{5}}{5}, \frac{6 \sqrt{5}}{5}$.
(Fuzhou City, Fujian Province, China High School Mathematics Competition, 1990)
[Solution] Let the volume of the tetrahedron be $V$, the areas of the four faces be $S_{1}, S_{2}, S_{3}, S_{4}$, and the heights from these faces be $h_{1}, h_{2}, h_{3}, h_{4}$. Thus,
$$
V=\frac{1}{3} S_{1} h_{1}=\frac{1}{3} S_{2} h_{2}=\frac{1}{3} S_{3} h_{3}=\frac{1}{3} S_{4} h_{4} \text {, }
$$

which means
$$
S_{1} h_{1}=S_{2} h_{2}=S_{3} h_{3}=S_{4} h_{4},
$$

so $S_{1}: S_{2}: S_{3}: S_{4}=\frac{1}{h_{1}}: \frac{1}{h_{2}}: \frac{1}{h_{3}}: \frac{1}{h_{4}}$.
For option $(B)$,
$$
\frac{1}{h_{1}}: \frac{1}{h_{2}}: \frac{1}{h_{3}}: \frac{1}{h_{4}}=\frac{1}{4}: \frac{\sqrt{3}}{25}: \frac{\sqrt{3}}{25}: \frac{\sqrt{3}}{25},
$$

and
$$
\frac{1}{4}=\frac{6}{24}>\frac{6}{25}>\frac{3 \sqrt{3}}{25} \text {, }
$$

which means if the above set of numbers were the heights of the tetrahedron, then one of its face areas would be greater than the sum of the areas of the other three faces, which is impossible. Therefore, the set of numbers in option $(B)$ cannot be the heights of the tetrahedron.
Hence, the answer is $(B)$.