Find all ordered triples $(x, y, z)$ of real numbers which satisfy the following system of equations:

$$
\left\{\begin{array}{l}
x y=z-x-y \\
x z=y-x-z \\
y z=x-y-z
\end{array}\right.
$$
Subtracting the second equation from the first gives \( x y - x z = 2 z - 2 y \). Factoring \( y - z \) from each side and rearranging gives

\[
(x + 2)(y - z) = 0
\]

so either \( x = -2 \) or \( z = y \).

If \( x = -2 \), the first equation becomes \( -2 y = z + 2 - y \), or \( y + z = -2 \). Substituting \( x = -2, y + z = -2 \) into the third equation gives \( y z = -2 - (-2) = 0 \). Hence either \( y \) or \( z \) is 0, so if \( x = -2 \), the only solutions are \( (-2, 0, -2) \) and \( (-2, -2, 0) \).

If \( z = y \) the first equation becomes \( x y = -x \), or \( x(y + 1) = 0 \). If \( x = 0 \) and \( z = y \), the third equation becomes \( y^2 = -2 y \) which gives \( y = 0 \) or \( y = -2 \). If \( y = -1 \) and \( z = y = -1 \), the third equation gives \( x = -1 \). So if \( y = z \), the only solutions are \( (0, 0, 0) \), \( (0, -2, -2) \) and \( (-1, -1, -1) \).

In summary, there are 5 solutions: \( (-2, 0, -2) \), \( (-2, -2, 0) \), \( (0, 0, 0) \), \( (0, -2, -2) \) and \( (-1, -1, -1) \).