Solution. We need to determine the digits $x, y, z$ such that

$$
\begin{aligned}
\overline{x y z} & =x+y+z+x y+y z+z x+x y z \\
100 x+10 y+z & =x+y+z+x y+y z+z x+x y z \\
99 x+9 y-x y & =y z+z x+x y z
\end{aligned}
$$

Therefore,

$$
z=\frac{99 x+9 y-x y}{x+y+x y}
$$

Since $z$ is a digit, it must be $z \leq 9$, i.e.,

$$
99 x+9 y-x y \leq 9 x+9 y+9 x y
$$

which implies $90 x \leq 10 x y$, or $y \geq 9$. Since $y$ is a digit, it must be $y=9$.

Substituting, we get $z=\frac{99 x+9 \cdot 9-9 x}{x+9+9 x}=\frac{90 x+81}{10 x+9}=9$.

The original equation is satisfied for $y=z=9$ and any $x$:

$$
x+y+z+x y+y z+z x+x y z=x+9+9+9 x+81+9 x+81 x=100 x+99=\overline{x 99}
$$

The sought numbers are: $199,299,399,499,599,699,799,899$ and 999.
1. Determine all three-digit numbers $\overline{x y z}$ (where $x, y, z$ are decimal digits) that are equal to the expression $x+y+z+x y+y z+z x+x y z$.