Solve the following system of equations:

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

First, let's find the solutions of the system of equations, if any, for which $x+y+z=0$. Under this condition, the system of equations transforms as follows:

$$
y^{2}=x z ; \quad z^{2}=x y ; \quad x^{2}=y z ; \quad \text { and } \quad x+y+z=0
$$

From the condition, we have $y=-x-z$, so $y^{2}=x^{2}+2 x z+z^{2}=x z$, which simplifies to $x^{2}+x z+z^{2}=0$. This can only be true for real numbers if $x=z=0$, from which it follows that $y=0$ as well. Thus, the only possible solution when $x+y+z=0$ is $x=y=z=0$, and this is indeed a solution to the system (1).

For the rest of the analysis, we assume that $x+y+z \neq 0$. Subtract the second equation from the first, the third from the second, and the first from the third. We then get the following equations:

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

After factoring and dividing by $x+y+z \neq 0$:

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

Let's choose the value of $x+y+z$ as a parameter and denote it by $3s$. Using this, we can express the values of $x, y$, and $z$ as follows:

$$
\begin{aligned}
& x+y+z=3s \\
& y-z=3(a-b)s \\
& z-x=3(b-c)s \\
& x-y=3(c-a)s
\end{aligned}
$$

By adding the first equation to the fourth and subtracting the third, we get the value of $x$. Similarly, we can find the values of $y$ and $z$ by adding or subtracting the appropriate equations:

$$
\begin{aligned}
& x=(2c-a-b+1)s \\
& y=(2a-b-c+1)s \\
& z=(2b-c-a+1)s
\end{aligned}
$$

Thus, if the system (1) has solutions other than $x=y=z=0$, they are given by (3). Let's check when (3) is a solution to the original system of equations (assuming $s \neq 0$). For example, substituting into the left side of the first equation, the obtained value is

$$
\left[\left(3a^{2}+3b^{2}+3c^{2}-3ab-3bc-3ca-3a-3b-3c\right)+9a\right] \cdot s^{2}
$$

while the right side is $9as^{2}$. The two sides will only be equal if

$$
a^{2}+b^{2}+c^{2}-(ab+bc+ca)-(a+b+c)=0
$$

It is easy to see that if condition (4) is satisfied, the other equations are also satisfied by (3). If condition (4) is satisfied, then (3) provides all solutions, including the solution $x=y=z=0$, which is obtained when $s=0$.

In summary, if (4) is not satisfied, the only solution to the system of equations is $x=y=z=0$. If (4) is true, then the system of equations has all solutions given by (3), where $s$ is any real number.