Let's determine all real triples $(a ; b ; c)$ for which the following system of equations has a solution different from $x=y=z=0$:

$$
\begin{aligned}
& a x+b y+c z=0 \\
& b x+c y+a z=0 \\
& c x+a y+b z=0
\end{aligned}
$$
I. solution. It is easily observable that if $a+b+c=0$, then every number triplet satisfying $x=y=z$ is a solution.

It is also clear that if $a=b=c$, then every number triplet satisfying $x+y+z=0$ is a solution.

In the following, we assume that $a+b+c \neq 0$, and at least two of $a, b, c$ are different, and finally, we assume that the system of equations has a solution $x_{1}, y_{1}, z_{1}$ where $z \neq 0$.

By introducing the numbers

$$
x^{\prime}=\frac{x_{1}}{z_{1}}, y^{\prime}=\frac{y_{1}}{z_{1}}
$$

we get:

$$
a x^{\prime}+b y^{\prime}=-c, \quad b x^{\prime}+c y^{\prime}=-a \text { and } c x^{\prime}+a y^{\prime}=-b
$$

Adding these three equations and dividing by $a+b+c \neq 0$, we arrive at the equation $x^{\prime}+y^{\prime}=-1$. Substituting $y^{\prime}=-1-x^{\prime}$ into our new three equations:

$$
(a-b) x^{\prime}=b-c, \quad(b-c) x^{\prime}=c-a \text { and }(c-a) x^{\prime}=a-b
$$

It is clear that no two of $a, b, c$ can be equal, i.e., if, for example, $a=b$, then from the first equation $b=c$, so $a=b=c$, which we have excluded. If we multiply these three equations, we get:

$$
(a-b)(b-c)(c-a)\left(x^{\prime 3}-1\right)=0
$$

Here, $(a-b)(b-c)(c-a) \neq 0$, so $x^{\prime 3}=1$, from which $x^{\prime}=1$, thus $a-b=b-c=c-a$. The sum of these three - equal - numbers is zero, so they can only be equal if each of them is zero, which we have excluded. We have arrived at a contradiction with our assumption. Therefore, the system of equations in the problem has a non-trivial solution (different from $x=y=z=0$) in two cases: when $a+b+c=0$ or when $a=b=c$.

II. solution. It is known that a homogeneous linear system of equations - as in the problem - has a non-trivial (non-zero) solution if and only if its determinant is zero. In our case, this determinant is:

$$
0=\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|=3 a b c-a^{3}-b^{3}-c^{3}=-\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]
$$

From this, we get two cases: Either $a+b+c=0$ (in which case every triplet $x=y=z$ is a solution), or $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$, i.e., $a=b=c$ (in which case every triplet satisfying $x+y+z=0$ is a solution). In all other cases, only the trivial solution exists.