12 、Place $0, 1, 2, 3, 4, 5, 6, 7$ on the eight vertices of a cube (each vertex has one number, and all numbers can only be used once), such that the sum of the two numbers on each edge is a prime number. Then the maximum sum of the four numbers on one face is
【Analysis】If the two numbers on a certain edge have the same parity, their sum must be even, so it can only be 2, meaning the two numbers on this edge can only be $0, 2$. Considering the face containing this edge, there are 4 numbers on this face, with $0, 2$ adjacent. Therefore, the other 2 numbers must be odd, and the sum of these 2 odd numbers is also even, but the minimum is $1+3=4$, which is definitely not a prime number.
Therefore, we should fill the numbers with alternating parity.
So, on one face, there should be 2 odd and 2 even numbers, with the maximum sum being $7+6+5+4=22$. Unfortunately, $4+5=9$ is a composite number, so it cannot be filled.
Since there are 2 odd and 2 even numbers, the sum must be even. Next, consider 20, with $7+6+5+2, 7+6+4+3$,

Unfortunately, $2+7, 3+6$ are both composite numbers, so it cannot be filled.
Now consider 18, there is a feasible way to fill it, so the maximum is 18.