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

$$
\left\{\begin{array}{l}
x^{2}-22 y-69 z+703=0 \\
y^{2}+23 x+23 z-1473=0 \\
z^{2}-63 x+66 y+2183=0
\end{array}\right.
$$
# 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:

$$
\left\{\begin{array}{l}
20^{2}-22 \cdot(-22)-69 \cdot 23+703=400+484-1587+703=0 \\
(-22)^{2}+23 \cdot 20+23 \cdot 23-1473=484+460+529-1473=0 \\
23^{2}-63 \cdot 20+66 \cdot(-22)+2183=529-1260-1452+2183=0
\end{array}\right.
$$

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