12. (6 points) The natural number $a$ is a multiple of 3, $a-1$ is a multiple of 4, $a-2$ is a multiple of 5, then the smallest $a$ is $\qquad$
【Analysis】Since the natural number $a$ is a multiple of 3, $a-1$ is a multiple of 4, and $a-2$ is a multiple of 5, then this natural number is divisible by 3, leaves a remainder of 1 when divided by 4, and leaves a remainder of 2 when divided by 5. The method of enumeration can be applied.

【Solution】Since the natural number $a$ is a multiple of 3, $a-1$ is a multiple of 4, and $a-2$ is a multiple of 5, so this number leaves a remainder of 2 when divided by 5.

Since the units digit of a multiple of 5 is 0 or 5, adding 2 results in a units digit of 2 or 7, and this number is also a multiple of 3, so we can list the multiples of 3 with a units digit of 2 or 7.

For example: 12, 27, 42, 57, .
And $(57-1) \div 4=14$,
So the number that meets the conditions is 57.
Therefore, the answer is: 57.