I don't understand this problem but i am still working on it....
But what is the right question? As is, its wording is extremely odd.
It seems to me the question is about line segments made of exactly four points taken from a lattice where the points are the 64 latice points, from .
We have three sets of four planes with ten lines in each plane.
BUT I don't know how we should count the diagonals.
In my interpretation there are 4 3D-diagonals, and a number more of 2D-diagonals.
Actually, I'm getting 61 lines.
(And you didn't answer my question.)
Edit: Hold on, I made a mistake counting...
Yep, I'm getting 76 lines now.
We can consider that as a plane parallel to the -plane. All coordinates look like .
There are ten of our lines in that plane: four parallel to the -axis; four parallel to the -axis; and two diagonals.
There are three copies of that plane, , all parallel to the -plane.
Now there are forty of our lines.
There are two more sets of those planes: 4 parallel to the -plane and 4 parallel to the -plane.
Now we are up to 120. We have 4 diagonals.
Hmm, it seems you counted 5 lines everywhere, where you should have counted 4.
It seems you included coordinate zero, but your problem statement says "positive" integers.
So for instance the diagonals you drew are actually 3x(4+4)=24.
Parallel to the axis you have 4x4 lines that intersect a face of your cube, which gives 3x4x4=64 lines.
Then your final 4 3D-lines gives 24+64+4=76.
Here's another way.
Suppose you represent each line by with .
Then must take the values 0,1,2,3 to make sure all coordinates are integers.
Then is either 1,2,3, or 4.
And can be zero in all cases.
If then can also be 1.
And if then can also be -1.
So there are 6 possibilities.
The same for y and z gives you possibilities.
But b cannot be the zero vector, because then you would not have distinct points.
So possibilities are wrong.
This gives you possibilities.
Since you count every line double, you need to divide this result by 2...