Solve the following system of equations, where $a, b$, and $c$ are real parameters:

$$
x y+y z+z x=a^{2}-x^{2}=b^{2}-y^{2}=c^{2}-z^{2}
$$
Add $x^{2}$ to the first two expressions, then - after factoring - we arrive at the equation

$$
(x+y)(x+z)=a^{2}
$$

Similarly, by adding $y^{2}$ to the first and third, and $z^{2}$ to the first and fourth, we get the equations

$$
\begin{aligned}
& (y+z)(y+x)=b^{2} \\
& (x+z)(y+z)=c^{2}
\end{aligned}
$$

1. Case: $a b c \neq 0$. Then by multiplying (1) and (2) and dividing by (3), we get

$$
(x+y)^{2}=\frac{a^{2} b^{2}}{c^{2}}
$$

and similarly, we obtain the equations

$$
(x+z)^{2}=\frac{a^{2} c^{2}}{b^{2}}, \quad(y+z)^{2}=\frac{b^{2} c^{2}}{a^{2}}
$$

From (1) and (2), it is clear that the signs of $x+y, x+z$, and $y+z$ are the same. Therefore, according to the sign of the sum of the unknowns, there are two possibilities:

$$
x+y=\left|\frac{a b}{c}\right|, \quad x+z=\left|\frac{a c}{b}\right|, \quad y+z=\left|\frac{b c}{a}\right|
$$

or

$$
x+y=-\left|\frac{a b}{c}\right|, \quad x+z=-\left|\frac{a c}{b}\right|, \quad y+z=-\left|\frac{b c}{a}\right|
$$

The solutions to these are

$$
\begin{aligned}
& x_{1}=\frac{\left|\frac{a b}{c}\right|+\left|\frac{a c}{b}\right|-\left|\frac{b c}{a}\right|}{2}, \quad y_{1}=\frac{\left|\frac{a b}{c}\right|-\left|\frac{a c}{b}\right|+\left|\frac{b c}{a}\right|}{2} \\
& z_{1}=\frac{-\left|\frac{a b}{c}\right|+\left|\frac{a c}{b}\right|+\left|\frac{b c}{a}\right|}{2}
\end{aligned}
$$

and $x_{2}=-x_{1}, y_{2}=-y_{1}, z_{2}=-z_{1}$.

Our steps are reversible, so in the case $a b c \neq 0$, these and only these are the solutions to the system of equations.

2. Case: $a=b=c=0$. Then at least two of $x+y, x+z, y+z$ must be zero, otherwise (1), (2), and (3) cannot all be satisfied simultaneously. If, for example, $x+y=0=x+z$, then from this $y=z=-x$ follows, and such triples of numbers satisfy the original system of equations as well.

Similarly, in the other two cases, we get the solutions $x=y=-z$ and $x=z=-y$.

3. Case: $a \neq 0, b=0$. Then from (2), $y+z=0$ or $y+x=0$. The latter is not possible, as it would imply $a=0$, so $y+z=0$. In this case, however, $c=0$ and $y=-z$. Then from (1), $a^{2}=x^{2}-y^{2}$, i.e., $x= \pm \sqrt{a^{2}+y^{2}}$. Thus, we get the solutions

$$
x= \pm \sqrt{a^{2}+y^{2}}, \quad y \text { arbitrary, } z=-y
$$

and these triples are also solutions to the original system of equations.

Similarly, in the case $a=b=0, c \neq 0$, we get the solutions $x$ arbitrary, $y=-x, z= \pm \sqrt{c^{2}+x^{2}}$, and in the case $a=c=0, b \neq 0$, we get the solutions $x$ arbitrary, $y= \pm \sqrt{b^{2}+x^{2}}, z=-x$.

It is obvious that the solutions of the 3rd case include the solutions of the 2nd case, so the solutions to the problem are briefly as follows:

If $a b c \neq 0$, then $x=x_{1}, y=y_{1}, z=z_{1}$ or $x=-x_{1}, y=-y_{1}, z=-z_{1}$.

If exactly one of $a, b, c$ is zero, then there is no solution.

If $a=b=0$, then $x$ is arbitrary, $y=-x$ and $z= \pm \sqrt{c^{2}+y^{2}}$.

If $a=c=0$, then $z$ is arbitrary, $x=-z$ and $y= \pm \sqrt{b^{2}+x^{2}}$.

If $b=c=0$, then $y$ is arbitrary, $z=-y$ and $x= \pm \sqrt{a^{2}+z^{2}}$.