9. In a tetrahedron, all plane angles at one vertex are right angles. A certain point in space is at a distance of 3 from the specified vertex of the tetrahedron, and at distances of $\sqrt{5}, \sqrt{6}$, and $\sqrt{7}$ from the other vertices. Find the distances from the center of the circumscribed sphere around the tetrahedron to each of its faces.

![](https://cdn.mathpix.com/cropped/2024_05_06_1326531d43e0d8faed41g-4.jpg?height=593&width=891&top_left_y=346&top_left_x=274)
Answer: $\frac{\sqrt{3}}{2}, 1, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{\sqrt{13}}$.

Let $ABCD$ be the given tetrahedron, and all plane angles at vertex $D$ are right angles. Let the lengths of segments $DA, DB$, and $DC$ be $a, b$, and $c$ respectively. We associate the rays $DA, DB$, and $DC$ with the rectangular coordinate system $Dxyz$. In this coordinate system, point $D$ has coordinates $(0,0,0)$, and points $A, B$, and $C$ have coordinates $(a, 0,0), (0, b, 0)$, and $(0,0, c)$ respectively. Let the coordinates of point $K$, specified in the problem, be $(x, y, z)$. Without loss of generality, assume that $KA=\sqrt{5}, KB=\sqrt{6}, KC=\sqrt{7}$. Then

\[
\begin{gathered}
KD^{2}=x^{2}+y^{2}+z^{2}=9 \\
KA^{2}=(x-a)^{2}+y^{2}+z^{2}=5 \\
KB^{2}=x^{2}+(y-b)^{2}+z^{2}=6 \\
KC^{2}=x^{2}+y^{2}+(z-c)^{2}=7
\end{gathered}
\]

Add the second, third, and fourth equations and subtract twice the first equation from the obtained sum:

\[
(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=0
\]

from which it follows that $x=a, y=b, z=c$, i.e., points $A, B, C$, and $K$ are vertices of a rectangular parallelepiped, and segment $KD$ is its main diagonal. Substituting the found values of $x, y$, and $z$ into equations (1)-(4), we find that $a^{2}=4, b^{2}=3, c^{2}=2$.

The center $O$ of the sphere circumscribed around the tetrahedron lies at the midpoint of segment $KD$. The distances from point $O$ to the faces $CDA, CDB$, and $DBA$ are $\frac{b}{2}, \frac{a}{2}$, and $\frac{c}{2}$ respectively, i.e., $\frac{\sqrt{3}}{2}, 1$, and $\frac{\sqrt{2}}{2}$.

Let $OH$ be the perpendicular dropped from point $O$ to the face $ABC$. We will compute the volume of the tetrahedron $OABC$ in two ways:

\[
V_{OABC}=\frac{1}{3} S_{ABC} \cdot OH=\frac{1}{3} S_{OAB} \cdot CF
\]

Find the area of $\triangle ABC$. By the cosine theorem, $\cos \angle ABC=\frac{BC^{2}+BA^{2}-AC^{2}}{2 BC \cdot BA}$. Since $BC=\sqrt{b^{2}+c^{2}}=\sqrt{5}, BA=\sqrt{a^{2}+b^{2}}=\sqrt{7}, AC=\sqrt{a^{2}+c^{2}}=\sqrt{6}$, then $\cos \angle ABC=\frac{5+7-6}{2 \sqrt{5} \sqrt{7}}=\frac{3}{\sqrt{35}}$. Therefore, $\sin \angle ABC=\frac{\sqrt{26}}{\sqrt{35}}$. Then

\[
S_{ABC}=\frac{1}{2} BA \cdot BC \sin \angle ABC=\frac{1}{2} \sqrt{7} \sqrt{5} \frac{\sqrt{26}}{\sqrt{35}}=\frac{\sqrt{26}}{2}
\]

From the right triangle $A_1B_1C$, we find $CF=\frac{A_1C \cdot B_1C}{A_1B_1}=\frac{ab}{\sqrt{a^{2}+b^{2}}}=\frac{2\sqrt{3}}{\sqrt{7}}$.
$S_{AOB}=\frac{1}{2} AB \cdot \frac{1}{2} AA_1=\frac{c \sqrt{a^{2}+b^{2}}}{4}=\frac{\sqrt{14}}{4}$. Finally, we get

\[
OH=\frac{CF \cdot S_{OAB}}{S_{ABC}}=\frac{2\sqrt{3}}{\sqrt{7}} \frac{\sqrt{14}}{4} \frac{2}{\sqrt{26}}=\frac{\sqrt{3}}{\sqrt{13}}
\]