jmc

Testing for divisibility by 9 does not work, because the sum of the digits is 18, so the digit could be either 0 or 9. Test for divisibility by 11. The alternating sum of the digits of $12!$ is $4-7+a-1+6=2+a$, which must be divisible by 11. Therefore, since $2+9=11$, we have $\boxed{a=9}$.