Solve the following system for real $x,y,z$ 
\[ \{ \begin{array}{ccc} x+ y -z & =& 4 \\ x^2 - y^2 + z^2 & = & -4 \\ xyz & =& 6. \end{array} \]
1. We start with the given system of equations:
   \[
   \begin{cases}
   x + y - z = 4 \\
   x^2 - y^2 + z^2 = -4 \\
   xyz = 6
   \end{cases}
   \]

2. From the first equation, we can express \(x\) in terms of \(y\) and \(z\):
   \[
   x = 4 - y + z
   \]

3. Substitute \(x = 4 - y + z\) into the second equation:
   \[
   (4 - y + z)^2 - y^2 + z^2 = -4
   \]

4. Expand and simplify the equation:
   \[
   (4 - y + z)^2 = 16 - 8y + y^2 + 8z - 2yz + z^2
   \]
   \[
   16 - 8y + y^2 + 8z - 2yz + z^2 - y^2 + z^2 = -4
   \]
   \[
   16 - 8y + 8z - 2yz + 2z^2 = -4
   \]
   \[
   2z^2 - 2yz + 8z - 8y + 16 = -4
   \]
   \[
   2z^2 - 2yz + 8z - 8y + 20 = 0
   \]

5. Simplify further:
   \[
   z^2 - yz + 4z - 4y + 10 = 0
   \]

6. Now, substitute \(x = 4 - y + z\) into the third equation:
   \[
   (4 - y + z)yz = 6
   \]
   \[
   (4yz - y^2z + yz^2) = 6
   \]

7. Notice that we have two equations involving \(y\) and \(z\). We can solve for \(y\) by isolating it in one of the equations. Let's solve the quadratic equation:
   \[
   y^2 - 4y - \frac{12}{y} = 16 + y^2 - 8y
   \]
   \[
   2y - \frac{3}{y} = 5
   \]

8. Multiply through by \(y\) to clear the fraction:
   \[
   2y^2 - 3 = 5y
   \]
   \[
   2y^2 - 5y - 3 = 0
   \]

9. Solve the quadratic equation using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   y = \frac{5 \pm \sqrt{25 + 24}}{4}
   \]
   \[
   y = \frac{5 \pm 7}{4}
   \]
   \[
   y = 3 \quad \text{or} \quad y = -\frac{1}{2}
   \]

10. For \(y = 3\):
    \[
    x = 1 + z \quad \text{or} \quad x = \frac{9}{2} + z
    \]
    \[
    xz = 2 \quad \text{or} \quad xz = -12
    \]

11. Solving for \(x = 1 + z\):
    \[
    (1 + z)z = 2
    \]
    \[
    z^2 + z - 2 = 0
    \]
    \[
    z = 1 \quad \text{or} \quad z = -2
    \]
    \[
    (x, y, z) = (2, 3, 1) \quad \text{or} \quad (x, y, z) = (-1, 3, -2)
    \]

12. For \(y = -\frac{1}{2}\), solving for \(x = \frac{9}{2} + z\) does not yield real solutions for \(z\).

Conclusion:
\[
(x, y, z) \in \{(2, 3, 1), (-1, 3, -2)\}
\]

The final answer is \(\boxed{(2, 3, 1), (-1, 3, -2)}\)