jmc

To get the largest number, first maximize the hundreds digit, then the tens, then the ones. One-digit factors of $12$ are $6$, $2$, $3$, $4$, and $1$, so the hundreds digit must be $6$. In order to make the digits multiply to 12, the next two digits must be $2$ and $1$. Therefore, the largest three-digit number in which the product of the digits is $12$ is $\boxed{621}$.