$100a + 10b + c = a! + b! + c!$
$$a! + b! + c! < 1000 \Rightarrow \max(a, b, c) \le 6 \\ (7! > 1000,\ 6! < 1000)$$
Because $a! + b! + c! > 99$, one of the digits $a$, $b$, or $c$ has to be equal to $5$ or $6$ (since $4! + 4! + 4! < 99$).
If one of the digits is equal to $6$, then $a! + b! + c! > 6! = 720$, which is a contradiction.
Now $\max(a, b, c) = 5$. We have three cases:

a) $5bc$
This case is not possible because the largest number is $5! + 5! + 5! = 360 < 500$.

b) $a5c$
The largest number in this case is $4! + 5! + 5! = 264$, so $a = 1$ or $a = 2$, i.e.
$$150,\ 1! + 5! + 0! \neq 150$$
$$151,\ 1! + 5! + 1! \neq 151$$
$$152,\ 1! + 5! + 2! \neq 152$$
$$153,\ 1! + 5! + 3! \neq 153$$
$$154,\ 1! + 5! + 4! \neq 154$$
$$155,\ 1! + 5! + 5! \neq 155$$
$$254,\ 2! + 5! + 4! < 200$$
$$255,\ 2! + 5! + 5! \neq 255$$

c) $ab5$
If $a = 1$ or $a = 2$, then the case is similar to b), i.e. solution is $1! + 4! + 5! = 145$.