$$
\begin{aligned}
x+y+z & =1 \\
1 / x+1 / y+1 / z & =1 \\
x y z & =a!
\end{aligned}
$$

The following system of equations needs to be solved:

$$
\begin{aligned}
x+y+z & =1 \\
1 / x+1 / y+1 / z & =1 \\
x y z & =a!
\end{aligned}
$$
II-b01: $\frac{1}{x}=1-\frac{1}{y}-\frac{1}{z}=\frac{y z-z-y}{y z}$ and $x=\frac{y z}{y z-z-y}$.

Substituting this into I: $\frac{y z}{y z-y-z}+y+z=1$.

Rearranging this:

$$
(y+z)[y z-(y+z)+1]=0
$$

Two cases are possible:

a.) $y+z=0$, in this case from I, $x=1$ and $y=-z$ leads to III $x y z=1$.

$(-z) z=-z^{2}=a, z= \pm \sqrt{-a}, y=\mp \sqrt{-a}, x=1$. This represents two solutions.

b.) $y z-(y+z)+1=0$.

Multiplying this by $x$ and considering I and III:

$$
x y z-x(y+z)+x=a-x(1-x)+x=a+x^{2}=0
$$

From here $x= \pm \sqrt{-a}$. Furthermore, $y+z=1-x, y z=\frac{a}{x}=\frac{a}{ \pm \sqrt{-a}}=-x$ so $y$ and $z$ are the roots of the following quadratic equation:

$$
{ }^{2}{ }^{2}-(1-\square)_{-}-\square=0
$$

From here $u=\frac{1-x \pm(1+x)}{2}, y=u_{1}=1, z=u_{2}=-x=\mp \sqrt{-a}$.

Thus: $\quad x= \pm \sqrt{-a}, \quad y=1, \quad z=\mp \sqrt{-a}$.

This is the same as case (a.), just the values of the unknowns are swapped. Since the unknowns are symmetric, the following is also a solution: $x= \pm \sqrt{-a}, y=\mp \sqrt{-a}, z=1$. Therefore, there are a total of six solutions.