Assigning numbers to edges and then adding them for the vertices

Question

Numbers $1, 2, 3, \ldots, 12$ are placed on all edges of a cube, one per edge. Then for each of the $8$ vertices of the cube, the sum of $3$ numbers converging on this vertex is computed. Is it possible all $8$ these sums are equal to each other?
I started by adding $1$ to $12$ up and then multiplying by two since each number is counted twice on the cube. However, I am not sure how to prove that there exists a configuration since my process involves the assumption that it works.

Answer 1

Adding $1$ to $12$ and getting $78$, then multiplying by $2$ to get $156$ gives us the total of all the vertices. If we then divide by $8$ we can find what each vertex has to equal. The fact that it is not an integer, being $\frac {39}2,$ says this is not possible.

Answer 2

The answer is no.

Note that the total weight at the $8$ corners are twice the sum of numbers $${\text {{1,2,3,4,5,6,7,8,9,10,11,12}}}$$

Twice the sum is $$\frac {2(12)(13)}{2} =156.$$

Since $156$ is not a multiple of $8$, there is no way that the edge labeling works as desired.