12. There are four distinct positive integers. Any two of them are taken to form a pair, and in the same pair, the larger number is subtracted by the smaller number, then the differences of each pair are added together, and the sum is exactly 18. If the product of these four numbers is 23100, find these four numbers.
Three, 12. Let's assume four different positive integers are $a_{1}, a_{2}, a_{3},$ $a_{4}\left(a_{1}>a_{2}>a_{3}>a_{4}\right)$.

According to the problem, any two numbers chosen from the four form a group, making a total of 6 groups. Thus,
$$
\begin{array}{l}
\left(a_{1}-a_{2}\right)+\left(a_{1}-a_{3}\right)+\left(a_{1}-a_{4}\right)+ \\
\left(a_{2}-a_{3}\right)+\left(a_{2}-a_{4}\right)+\left(a_{3}-a_{4}\right)=18,
\end{array}
$$

which simplifies to $3\left(a_{1}-a_{4}\right)+\left(a_{2}-a_{3}\right)=18$.
Since $3\left(a_{1}-a_{4}\right)$ and 18 are both multiples of 3, $a_{2}-a_{3}$ must also be a multiple of 3.
Given the assumption $a_{1}>a_{2}>a_{3}>a_{4}$, we have $a_{2}-a_{3}a_{2}>a_{3}>a_{4}$, which implies
$$
a_{1}-a_{2}=1, a_{3}-a_{4}=1 \text {. }
$$

Also, $a_{1} a_{2} a_{3} a_{4}=23100=2 \times 2 \times 3 \times 5 \times 5 \times 7 \times 11$,
Through trial, we find $a_{1}=15, a_{2}=14, a_{3}=11, a_{4}=10$.
Therefore, the four numbers are $10,11,14,15$.