Find the height of a triangular pyramid, the lateral edges of which are pairwise perpendicular and equal to 2, 3, and 4.

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Let the lateral edges $DA, DB$, and $DC$ of the triangular pyramid $ABCD$ be pairwise perpendicular, and $DA=2, DB=3$, $DC=4$. Consider the triangular pyramid $ABCD$ with vertex $C$ and base $ABD$. Its lateral edge $CD$ is perpendicular to the two intersecting lines $BD$ and $AD$ in the plane $ABD$. Therefore, $CD$ is perpendicular to this plane. Thus, $CD=4$ is the height of the pyramid $ABCD$. The base of this pyramid is a right triangle $ABD$ with legs $AD=2$ and $BD=3$. Therefore,

$$
V_{ABCD}=\frac{1}{3} S_{\triangle ABD} \cdot CD=\frac{1}{3} \cdot \frac{1}{2} AD \cdot BD \cdot CD=\frac{1}{6} \cdot 2 \cdot 3 \cdot 4=4.
$$

Let $DK$ be the desired height of the pyramid $ABCD$. Then

$$
V_{ABCD}=\frac{1}{3} S_{\triangle ABC} \cdot DK
$$

From this, we find that

$$
DK=\frac{3 V_{ABCD}}{S_{\triangle ABC}}=\frac{12}{S_{\triangle ABC}}.
$$

It remains to find the area of triangle $ABC$. Let the lines $AK$ and $BC$ intersect at point $M$. Then the line $BC$ is perpendicular to the two intersecting lines $DK$ and $AD$ in the plane $AMD$. Therefore, $BC \perp DM$, i.e., $DM$ is the height of the right triangle $BCD$ drawn from the vertex of the right angle, and $AM$ is the height of triangle $ABC$. Further, we have:

$$
\begin{gathered}
DM=\frac{CD \cdot BD}{BC}=\frac{4 \cdot 3}{\sqrt{4^2+3^2}}=\frac{12}{5} \\
AM=\sqrt{DM^2 + AD^2}=\sqrt{\left(\frac{12}{5}\right)^2 + 4}=\frac{2\sqrt{61}}{5} \\
S_{\triangle ABC}=\frac{1}{2} BC \cdot AM=\frac{1}{2} \cdot 5 \cdot \frac{2\sqrt{61}}{5}=\sqrt{61}
\end{gathered}
$$

Therefore,

$$
DK=\frac{12}{S_{\triangle ABC}}=\frac{12}{\sqrt{61}}.
$$

## Answer

$\frac{12}{\sqrt{61}}$.