To find all triples $(a, b, c)$ of real numbers satisfying the system of equations:

$$\begin{align*} 1. \quad & a^2 + ab + c = 0, \\ 2. \quad & b^2 + bc + a = 0, \\ 3. \quad & c^2 + ca + b = 0, \end{align*}$$

we proceed by considering possible symmetric solutions first and also evaluate the possibility of all variables being zero.

### Step 1: Checking the trivial solution $(0, 0, 0)$

Substituting $a = 0$, $b = 0$, $c = 0$ into the equations, we find:

$$\begin{align*} 0^2 + 0 \cdot 0 + 0 &= 0, \\ 0^2 + 0 \cdot 0 + 0 &= 0, \\ 0^2 + 0 \cdot 0 + 0 &= 0. \end{align*}$$

Since all the equations are satisfied, $(0, 0, 0)$ is a valid solution.

### Step 2: Considering symmetric solutions where $a = b = c = x$

Assume $a = b = c = x$. Then the equations become:

$$\begin{align*} x^2 + x \cdot x + x &= 0 \quad \Rightarrow \quad 2x^2 + x = 0, \\ x^2 + x \cdot x + x &= 0 \quad \Rightarrow \quad 2x^2 + x = 0, \\ x^2 + x \cdot x + x &= 0 \quad \Rightarrow \quad 2x^2 + x = 0. \end{align*}$$

Each equation simplifies to:

$$2x^2 + x = 0$$

Factoring gives:

$$x(2x + 1) = 0$$

Thus, the solutions for $x$ are $x = 0$ or $x = -\frac{1}{2}$.

- For $x = 0$, this corresponds to the solution $(0, 0, 0)$.
- For $x = -\frac{1}{2}$, this gives the solution $\left( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} \right)$.

Both solutions satisfy the original equations, since substituting back verifies equality:

$$\begin{align*} \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) + \left(-\frac{1}{2}\right) &= \frac{1}{4} + \frac{1}{4} - \frac{1}{2} = 0, \\ \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) + \left(-\frac{1}{2}\right) &= 0, \\ \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) + \left(-\frac{1}{2}\right) &= 0. \end{align*}$$

### Final Solutions

Thus, the complete set of solutions for $(a, b, c)$ is:

$$\boxed{(0, 0, 0) \quad \text{and} \quad \left(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\right)}$$