2. Find all solutions to the system of 3 equations with 3 unknowns $x_{1}, x_{2}, x_{3}$ :

$$
\left\{\begin{array}{l}
x_{1}\left|x_{1}\right|-\left(x_{1}-a\right)\left|x_{1}-a\right|=x_{2}\left|x_{2}\right| \\
x_{2}\left|x_{2}\right|-\left(x_{2}-a\right)\left|x_{2}-a\right|=x_{3}\left|x_{3}\right| \\
x_{3}\left|x_{3}\right|-\left(x_{3}-a\right)\left|x_{3}-a\right|=x_{1}\left|x_{1}\right|
\end{array}\right.
$$

where \(a\) is a given real number.
Solution. Note that if $\left(x_{1}^{\circ}, x_{2}^{\circ}, x_{3}^{\circ}\right)$ is a solution to the system with parameter $a$, then $\left(-x_{1}^{\circ},-x_{2}^{\circ},-x_{3}^{\circ}\right)$ will be a solution with parameter $-a$. Therefore, it is sufficient to consider only the case when $a \geq 0$.

Let $(x_{1}, x_{2}, x_{3})$ be a solution for which $x_{1} \leq a, x_{2} \leq a, x_{3} \leq a$; then

$$
\left|x_{i}-a\right|=-\left(x_{i}-a\right)
$$

and if we add the equations, we get

$$
\left(x_{1}-a\right)^{2}+\left(x_{2}-a\right)^{2}+\left(x_{3}-a\right)^{2}=0
$$

from which it follows that $x_{1}=x_{2}=x_{3}=a$ is the only solution.

Let $(x_{1}, x_{2}, x_{3})$ be a solution for which $x_{i} \geq a$ for some $i$. Then from the $i$-th equation we get

$$
x_{i+1}\left|x_{i+1}\right|=x_{i}^{2}-\left(x_{i}-a\right)^{2}=a\left(2 x_{i}-a\right) \geq a^{2}
$$

Thus, $x_{i} \geq a$. For $i=3$, replace $i+1$ with 1; then we get $x_{1} \geq a, x_{2} \geq a$, $x_{3} \geq a$, and by adding the equations we get

$$
x_{1}=x_{2}=x_{3}=a
$$

Therefore, the only solution is $x_{1}=x_{2}=x_{3}=a$.