Find all possible real values of $a$ for which the system of equations
\[\{\begin{array}{cc}x +y +z=0\\\text{ } \\ xy+yz+azx=0\end{array}\]
has exactly one solution.
1. We start with the given system of equations:
   \[
   \begin{cases}
   x + y + z = 0 \\
   xy + yz + azx = 0
   \end{cases}
   \]
   Notice that \((x, y, z) = (0, 0, 0)\) is always a solution. We need to find the values of \(a\) for which this is the only solution.

2. Substitute \(y = -(x + z)\) from the first equation into the second equation:
   \[
   x(-(x + z)) + (-(x + z))z + azx = 0
   \]
   Simplify the equation:
   \[
   -x^2 - xz - z^2 + azx = 0
   \]
   Rearrange the terms:
   \[
   x^2 - xz(a - 2) + z^2 = 0
   \]

3. Let \(t = \frac{x}{z}\) where \(z \neq 0\). Then the equation becomes:
   \[
   t^2 - t(a - 2) + 1 = 0
   \]

4. For \((0, 0, 0)\) to be the only solution, the quadratic equation \(t^2 - t(a - 2) + 1 = 0\) must have no real solutions. This means the discriminant \(\Delta\) must be less than 0:
   \[
   \Delta = (a - 2)^2 - 4 < 0
   \]

5. Solve the inequality:
   \[
   (a - 2)^2 - 4 < 0
   \]
   \[
   (a - 2)^2 < 4
   \]
   \[
   -2 < a - 2 < 2
   \]
   Add 2 to all parts of the inequality:
   \[
   0 < a < 4
   \]

6. Therefore, the values of \(a\) for which the system has exactly one solution \((0, 0, 0)\) are in the interval \((0, 4)\).

The final answer is \( \boxed{ 0 < a < 4 } \)