12. Select 3 cards from five number cards $0,4,6,7,8$ to form a two-digit number, then a total of $\qquad$ different three-digit numbers can be formed (Note: Cards can be flipped).
【Analysis】Method one:
I. If both 0 and 6 are selected, there are 3 ways to choose the cards. In this case, the 3 cards have $2 \times 2 \times 1=4$ arrangements (0 cannot be the first digit), and 6 can be flipped to become 9, so a total of $3 \times 4 \times 2=24$ three-digit numbers can be formed.
II. If 6 is selected but 0 is not, there are 3 ways to choose the cards. In this case, the 3 cards have $3 \times 2 \times 1=6$ arrangements, and 6 can be flipped to become 9, so a total of $3 \times 6 \times 2=36$ three-digit numbers can be formed.
III. If 6 is not selected, then there are four remaining cards. The hundreds place has 3 choices (0 cannot be the first digit), the tens place has 3 choices, and the units place has 2 choices, so a total of $3 \times 3 \times 2=18$ three-digit numbers can be formed.
In summary, a total of 78 three-digit numbers can be formed.
Method two.
Considering 6 as 6, then $0, 4, 6, 7, 8$ can form $4 \times 4 \times 3=48$ three-digit numbers.
Considering 6 as 9, then $0, 4, 9, 7, 8$ can form $4 \times 4 \times 3=48$ three-digit numbers.
Considering the repeated numbers are those without 6 or 9, i.e., those formed by $0, 4, 7, 8$, there are $3 \times 3 \times 2=18$ three-digit numbers.
In summary, there are $48+48-18=78$ three-digit numbers.