$\overline{xyz} =x!+y!+z!$ how to find all of 3-digit integer number $\overline{xyz}$  such that:
$\overline{xyz} =x!+y!+z!$

As I was partway through typing this answer, Dan Shved's answer appeared which is much better than mine, but I'll include what I have so far as I use some different ideas. I reduce the number of possibilities to $13$ (though of course, I could reduce this number even further using the reasons in Dan's answer).

\begin{align*}
0! &= 1\\
1! &= 1\\
2! &= 2\\
3! &= 6\\
4! &= 24\\
5! &= 120\\
6! &= 720
\end{align*}
As $7! = 5040$, each of $x, y,$ and $z$ is between $0$ and $6$. Furthermore, at most one of $x, y,$ and $z$ can be $6$ otherwise $x! + y! + z!$ is not a three digit number (too big). Similarly, at least one of $x, y,$ and $z$ is a $5$ or $6$ otherwise $x! + y! + z!$ is not a three digit number (too small). Also, $x$ can't be $0$. We are left with $229$ possibilities.
Note that $\overline{xyz}$ can't end with a $0$ or a $1$ because there is no possible $x$ and $y$ such that $x! + y! \equiv 9\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $2$ or a $4$, we must have $\overline{xy} \in \{34, 43, 56, 65\}$
as these are the only choices with $x! + y! \equiv 0\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $3$ we must have $\overline{xy} \in \{13, 31\}$  as these are the only choices with $x! + y! \equiv 7\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $5$ we must have $\overline{xy} \in \{14, 40, 41\}$  as these are the only choices with $x! + y! \equiv 5\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $6$ we must have $\overline{xy} \in \{24, 42\}$ as these are the only choices with $x! + y! \equiv 6\ \textrm{mod} 10$.
So we are left with $15$ possibilities: 
$$342, 432, 562, 652, 344, 434, 564, 654, 133, 313, 145, 405, 415, 246, 426.$$
Note that $\overline{xyz}$ is divisible by $\min\{x!, y!, z!\} = (\min\{x, y, z\})!$. This condition is automatically satisfied if the smallest digit is $0, 1,$ or $2$. However, this condition rules out both $564$ and $654$ as they're not divisible by $4!$.
