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

$$
\left\{\begin{array}{l}
x^{2}-23 y+66 z+612=0 \\
y^{2}+62 x-20 z+296=0 \\
z^{2}-22 x+67 y+505=0
\end{array}\right.
$$
# Solution.

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

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

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

$$
\left\{\begin{array}{l}
(-20)^{2}-23 \cdot(-22)+66 \cdot(-23)+612=400+506-1518+612=0 \\
(-22)^{2}+62 \cdot(-20)-20 \cdot(-23)+296=484-1240+460+296=0 \\
(-23)^{2}-22 \cdot(-20)+67 \cdot(-22)+505=529+440-1474+505=0
\end{array}\right.
$$

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