[Median of a pyramid (tetrahedron) $)]$ [Median of a pyramid (tetrahedron) ]

Given a tetrahedron $A B C D$. All plane angles at vertex $D$ are right angles; $D A=1, D B=2, D C=3$. Find the median of the tetrahedron, drawn from vertex $D$.
Let $M$ be the point of intersection of the medians of triangle $ABC$. Then, therefore

$$
\begin{gathered}
\overrightarrow{DM2} 2=\frac{1}{9}(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}) 2= \\
=\frac{1}{9}(DA 2+DB 2+DC 2+2 \overrightarrow{DA} \cdot \overrightarrow{DB}+2 \overrightarrow{DA} \cdot \overrightarrow{DC}+2 \overrightarrow{DB} \cdot \overrightarrow{DC})= \\
=\frac{1}{9}(DA 2+DB 2+DC 2+0+0+0)=\frac{1}{9}(1+4+9)=\frac{14}{9} .
\end{gathered}
$$

Therefore, $DM=\frac{\sqrt{\mathbf{14}}}{3}$.

## Answer

$\frac{\sqrt{14}}{3}$.