5. Drop perpendiculars $D D_{1}, D D_{2}, D D_{3}$ from point $D$ to the planes $S B C$, $S A C$, and $S A B$ respectively. Let $D D_{1}=x, D D_{2}=y, D D_{3}=z$. According to the condition, we form the system of equations

$$
\left\{\begin{array}{l}
y^{2}+z^{2}=5 \\
x^{2}+z^{2}=13 \\
x^{2}+y^{2}=10
\end{array}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_778d240aea7dd323243eg-9.jpg?height=526&width=630&top_left_y=95&top_left_x=1848)

Fig. 6:

From here, we find $x=3, y=1, z=2$. Let the lengths of the edges $S A, S B$, and $S C$ be $a, b$, and $c$ respectively. Since points $A, B, C$, and $D$ lie in the same plane, the relation $\frac{3}{a}+\frac{1}{b}+\frac{2}{c}=1$ holds.

Using the inequality between the arithmetic mean and the geometric mean for three variables, we get:

$$
\begin{aligned}
& \frac{\frac{3}{a}+\frac{1}{b}+\frac{2}{c}}{3} \geqslant \sqrt[3]{\frac{3}{a} \cdot \frac{1}{b} \cdot \frac{2}{c}}=\sqrt[3]{\frac{6}{a b c}} \Longleftrightarrow \\
& \Longleftrightarrow 1=\left(\frac{3}{a}+\frac{1}{b}+\frac{2}{c}\right)^{3} \geqslant \frac{6 \cdot 27}{a b c} \Longleftrightarrow a b c \geqslant 6 \cdot 27
\end{aligned}
$$

with equality holding when $\frac{3}{a}=\frac{1}{b}=\frac{2}{c}=\frac{1}{3}$. The volume of the pyramid $V=\frac{a b c}{6}$, so $V \geqslant 27$. Equality holds when $a=9, b=3, c=6$.
Answer: 27.

Answer to option 17-2: 108.

Answer to option $17-3: 27$.

Answer to option $17-4: 108$.


[^0]:    ${ }^{1}$ This equality can be proven by expressing $B C^{2}$ from two triangles $B A C$ and $B D C$ using the planimetric cosine theorem.