3-ча 1. Solve the system of equations:

$$
\left\{\begin{aligned}
x^{2}+y^{2}-2 z^{2} & =2 a^{2} \\
x+y+2 z & =4\left(a^{2}+1\right) \\
z^{2}-x y & =a^{2}
\end{aligned}\right.
$$
Solve 1. Let's write these equations as follows:

$$
\left\{\begin{aligned}
x^{2}+y^{2} & =2 z^{2}+2 a^{2} \\
x+y & =4\left(a^{2}+1\right)-2 z \\
-xy & =a^{2}-z^{2}
\end{aligned}\right.
$$

Square the second equation, add to it the third equation multiplied by 2, and subtract the first equation. The result is:

$$
0=16\left(a^{2}+1\right)^{2}-16\left(a^{2}+1\right) z
$$

i.e., $z=a^{2}+1$. Now the second and third equations can be written as:

$$
\left\{\begin{aligned}
x+y & =2\left(a^{2}+1\right) \\
xy & =a^{4}+a^{2}+1
\end{aligned}\right.
$$

Solving this system reduces to solving a quadratic equation; solving it, we find

$$
x=a^{2} \pm a+1, \quad y=a^{2} \mp a+1
$$