Task 3. (15 points) Solve the system of equations

$$
\left\{\begin{array}{l}
x^{2}-23 y-25 z=-681 \\
y^{2}-21 x-21 z=-419 \\
z^{2}-19 x-21 y=-313
\end{array}\right.
$$
# Solution.

Add the first equation to the other two and complete the square for each variable:

$$
\begin{aligned}
& x^{2}+y^{2}+z^{2}-40 x-44 y-46 z=-1413 \\
& x^{2}+y^{2}+z^{2}-40 x-44 y-46 z+1413=0 \\
& x^{2}-40 x+400+y^{2}-44 y+484+z^{2}-46 z+529=0 \\
& (x-20)^{2}+(y-22)^{2}+(z-23)^{2}=0
\end{aligned}
$$

Therefore, \( x=20, y=22, z=23 \) is the only possible solution. Let's verify this by substituting into the system of equations:

$$
\left\{\begin{array}{l}
20^{2}-23 \cdot 22-25 \cdot 23=400-506-575=-681 \\
22^{2}-21 \cdot 20-21 \cdot 23=484-420-483=-419 \\
23^{2}-19 \cdot 20-21 \cdot 22=529-380-462=-313
\end{array}\right.
$$

Answer. \((20 ; 22 ; 23)\).