111. In the pyramid $S A B C$, the products of the lengths of the edges of each of the four faces are equal to the same number. The length of the height of the pyramid, dropped from $S$ to the face $A B C$, is $2 \sqrt{\frac{102}{55}}$, and the measure of the angle $C A B$ is $\arccos \left(\frac{1}{6} \sqrt{\frac{17}{2}}\right)$. Find the volume of the pyramid $S A B C$, if

$$
|S A|^{2}+|S B|^{2}-5|S C|^{2}=60
$$
111. Equality of the products of the lengths of the edges at each face means that the opposite edges of the pyramid are equal. Let's complete the pyramid $S A B C$ in the usual way to form a parallelepiped by drawing a plane through each edge parallel to the opposite edge. Due to the equality of the opposite edges of the pyramid $S A B C$, the resulting parallelepiped will be rectangular. Denote the edges of this parallelepiped as $a, b$, and $c$, as shown in Fig. 18.

Draw the height $D L$ in $\triangle E C D$. From $\triangle B C D$, we find

$$
\begin{gathered}
|D L|=\frac{b c}{\sqrt{\bar{b}^{2}+c^{2}}} \\
|A L|=\sqrt{a^{2}+|D L|^{2}}=\frac{\sqrt{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}}}{\sqrt{\overline{b^{2}+c^{2}}}} \\
S_{A B C}=\frac{1}{2} \sqrt{\overline{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}}}
\end{gathered}
$$

The volume of the pyramid $S A B C$ is $1 / 3$ of the volume of the parallelepiped. Given the height to the face $A B C$, we obtain the equation

$\sqrt{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}} \cdot \sqrt{\frac{102}{55}}=a b c$.

Using the cosine theorem for $\triangle A B C$, we get

$$
\begin{aligned}
& 6 a^{2}= \\
& =\sqrt{a^{2}+c^{2}} \cdot \sqrt{a^{2}+b^{2}} \cdot \sqrt{\frac{17}{2}} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-072.jpg?height=304&width=425&top_left_y=506&top_left_x=652)

Fig: 18.

Finally, the last condition of the problem gives us

$$
c^{2}-2 a^{2}-2 b^{2}=30
$$

Solving the system (1)-(3), we find $a^{2}=34, b^{2}=2, c^{2}=102$.

$$
\text { Answer: } \frac{34 \sqrt{6}}{3} \text {. }
$$