12. (3 points) The figure is a regular hexagon with a side length of 1, divided into six small triangles. Fill the numbers 4, 6, 8, 10, 12, 14, and 16 into the 7 circles. Adjacent pairs of small triangles form 6 rhombuses. Add the numbers at the four vertices of each rhombus and fill the sum in the center of the rhombus at positions $A, B, C, D, E, F$ (for example: $a+b+g+f=A$). It is known that $A, B, C, D, E, F$ are divisible by 2, 3, 4, 5, 6, 7 respectively. Therefore, $a \times g \times d=$ . $\qquad$
Given that the positions of the two 0s are 6 and 12, the multiple of 7 is $14+8=22$, leaving the numbers $4, 10, 16$ which have a remainder of 1 when divided by 3.

Only $14+8+4+16=42$ is a multiple of 7. So $d=10$.
The sum of $b$ and $c$ is 18, so $12+6+10+14=42$ is not a multiple of 4, thus the middle number is $g=8, f=14$.
According to the multiple of 5, the sum of $10+8$ has a remainder of 3 when divided by 5, so the sum of $c+e$ has a remainder of 2 when divided by 5.
【Solution】Solution: First consider the sum of the vertices of the rhombus being a multiple of $3$ and $6$. The remainders when the 7 numbers are divided by 3 are $1$, $0$, $2$, $1$, $0$, $2$, $1$.
A multiple of 6 is also a multiple of 3, and the combination that is a multiple of 6 is $1,1,2,2$ four numbers combined, so the combination that is a multiple of 3 is $0,1,0,2$ four numbers combined.

The middle number has a remainder of 2, so the numbers 8 and 14 have a remainder of 2 when divided by 3. Therefore, the middle number $g=8$ or 14,
$12+4=16$ (not satisfied), $12+16=28$ (not satisfied), $6+4=10$ (not satisfied), $6+16=22$ (satisfies the condition)
$$
a \times g \times d=4 \times 8 \times 10=320 \text {; }
$$

Therefore, the answer is: 320.