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

$$
\left\{\begin{array}{l}
x^{2}+25 y+19 z=-471 \\
y^{2}+23 x+21 z=-397 \\
z^{2}+21 x+21 y=-545
\end{array}\right.
$$
# Solution.

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

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

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

$$
\left\{\begin{array}{l}
(-22)^{2}+25 \cdot(-23)+19 \cdot(-20)=484-575-380=-471 \\
(-23)^{2}+23 \cdot(-22)+21 \cdot(-20)=529-506-420=-397 \\
(-20)^{2}+21 \cdot(-22)+21 \cdot(-23)=400-462-483=-545
\end{array}\right.
$$

Answer. \((-22, -23, -20)\).