12. (10 points) Among three-digit numbers, some numbers are 19 times the sum of their digits, such as $190=19 \times(1+9+0)$, please write down all such three-digit numbers.
【Analysis】First, this number is a multiple of 19, and it is a three-digit number. The sum of the digits of a three-digit number does not exceed 27, so this number does not exceed $19 \times 27=513$, which is more than 500. The sum of the digits does not exceed $4+9+9=22$, meaning this number does not exceed $22 \times 19=418$. We need to find all multiples of 19 within 418 and check if they meet the conditions to solve the problem.
【Solution】Solution: This number is a multiple of 19, and the sum of the digits of a three-digit number does not exceed 27, so this number does not exceed $19 \times 27=513$, which is more than 500. The sum of the digits does not exceed $4+9+9=22$, meaning this number does not exceed $22 \times 19=418$,
$$
\begin{array}{l}
19 \times 6=114 \quad(\sqrt{ }) 1+1+4=6 \\
19 \times 7=133 \quad(\sqrt{ }) 1+3+3=7 \\
19 \times 8=152 \quad(\sqrt{ }) 1+5+2=8 \\
19 \times 9=171 \quad(\sqrt{ }) 1+7+1=9 \\
19 \times 10=190 \quad(\sqrt{ }) 1+9+0=10 \\
19 \times 11=209 \quad(\sqrt{ }) 2+0+9=11 \\
19 \times 12=228 \quad(\sqrt{ }) 2+2+8=12 \\
19 \times 13=247 \quad(\sqrt{ }) 2+4+7=13 \\
19 \times 14=266 \quad(\sqrt{ }) \cdots \\
19 \times 15=285 \quad(\sqrt{ }) \cdots \\
19 \times 16=304 \quad 3+0+4=7 \neq 16 \\
19 \times 17=323 \quad \cdots \\
19 \times 18=342 \quad \cdots \\
19 \times 19=361 \quad \cdots \\
19 \times 20=380 \quad \cdots \\
19 \times 21=399 \quad(\sqrt{ }) 3+9+9=21 \\
19 \times 22=418 \quad 4+1+8=13 \neq 22 \\
\end{array}
$$

So all the results are:
114, 133, 152, 171, 190, 209, 228, 247, 266, 285, 399, a total of 11 numbers,
Answer: The three-digit numbers that meet the conditions are $114, 133, 152, 171, 190, 209, 228, 247, 266, 285, 399$.