353. Solve the systems of equations:
a) $\left\{\begin{array}{l}2 x=(y+z)^{2}, \\ 2 y=(z+x)^{2}, \\ 2 z=(x+y)^{2}\end{array}\right.$,
b) $\left\{\begin{array}{l}x^{2}-x y-x z+z^{2}=0, \\ x^{2}-x z-y z+3 y^{2}=2, \\ y^{2}+x y+y z-z^{2}=2 .\end{array}\right.$
$\triangle$ a) From the equations of the system, it is clear that the unknowns $x, y$ and $z$ must be non-negative.

Let's try using inequalities. Could it be that $x > y$?

Assume that $x > y$. Then from the first two equations, it follows that

$$
(y+z)^{2} > (z+x)^{2}, \quad y+z > z+x, \quad y > x
$$

But the inequalities $x > y$ and $y > x$ are incompatible. The only remaining possibility is $x = y$.

Similarly, the impossibility of the inequalities

$$
x < z, \quad x > z, \quad y < z
$$

can be proven. Thus, $x = y = z$. Further solution is straightforward.

b) Add the first two equations of the system and subtract the third. Why? To obtain the terms $-2xy, -2xz, -2yz$ and group them together with $x^2, y^2, z^2$ into squares of differences. We will have:

$$
\begin{gathered}
2x^2 - 2xy + 2y^2 + 2z^2 - 2xz - 2yz = 0 \\
(x-y)^2 + (y-z)^2 + (x-z)^2 = 0
\end{gathered}
$$

The obtained sum of squares is non-negative, and it equals zero only when each square is zero. Hence,

$$
x - y = 0, \quad y - z = 0, \quad x - z = 0 \Rightarrow x = y = z
$$

The subsequent solution is also simple.

Answer: a) $(0; 0; 0), (1/2; 1/2; 1/2)$, b) $(1; 1; 1), (-1; -1; -1)$.