Solve the system

$$
\left\{\begin{array}{l}
a y+b x=c \\
c x+a z=b \\
b z+c y=a
\end{array}\right.
$$

What is the geometric meaning of this system?
1) Let $a b c \neq 0$. Multiply the first equation by $c$, add the second, multiplied by $b$, and subtract the third, multiplied by $a$. We get

$2 b c x=c^{2}+b^{2}-a^{2}$. Similarly, $y$ and $z$ are found.

2) If $a=b=0$, then $c=0$ as well, and $x, y, z$ are any numbers.
3) If $a=0, b, c \neq 0$, then $b x=c,\left(b^{2}-c^{2}\right) x=0$, hence, when $b= \pm c \quad x= \pm 1$, and in other cases, there are no solutions. From the last equation

$y \pm z=0$.

## Answer

When $a b c \neq 0$, $x=\frac{b^{2}+c^{2}-a^{2}}{2 b c}, y=\frac{a^{2}+c^{2}-b^{2}}{2 a c}, z=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$.

When $a=0$, if $b=c \neq 0$, then $(x, y, z)=(1, t,-t)$; if $b=-c \neq 0$, then $(x, y, z)=(1, t, t)$.

When $b=0$, if $a=c \neq 0$, then $(x, y, z)=(t, 1,-t)$; if $a=-c \neq 0$, then $(x, y, z)=(t, 1, t)$.

When $c=0$, if $a=b \neq 0$, then $(x, y, z)=(t,-t, 1)$; if $a=-b \neq 0$, then $(x, y, z)=(t, t, 1)$.

When $a=b=c=0$, $x, y, z$ are any numbers.

In all other cases, there are no solutions.