12. Using the digits $1,2,3, \cdots 9$, without repetition, form 3 three-digit numbers such that the sum of the first two equals the third, for example: $245+736=981$. The minimum value of the sum of these three three-digit numbers is $\qquad$
12. 918 .
$173+286=459$
$1+2+3+\cdots+9=45$ is an odd number.
There is exactly 1 carry in the addition of three-digit numbers, so the sum of the 3 digits of the result is 18.
If it is $4 \overline{a b}$, then $a+b=14$, (so $\overline{4 a b} \geqslant 459$.)
If the sum is $\overline{3 a b}$, then $a+b=15$, let $\overline{c d e}+\overline{f q h}=\overline{3 a b}$
then $d+q=10+a$ or $e+h=10+b$. If the former, then $d+q \geqslant 16$ (because $10+a$ $\geqslant 1+a+b)$.
$\Rightarrow \max (d, q)=9$ (because $d \neq q$ )
So $a \geqslant 15-8=7$.
So $d+q \geqslant 17=9+8$.
$\Rightarrow(d, q)=(8,9), a=7$ then $b=8=d$
Contradiction.
The latter case is similarly unsolvable.

If the sum is $\overline{1 a b}$ or $\overline{2 a b}$, it is easy to verify that there is no solution by the same reasoning.