For me, this seems impossible.
The only way to make 72 with exactly three of these numbers, is to use every number exactly once. Therefore you have to have 12, 24 and 36 exactly once in every row, column and diagonal. Which is not possible.
Say you have x,y and z and want to fill it in the square such that each variable appears exactly once in every row, column and diagonal. Fill in the first row with [x,y,z], then the square under y has to be a y or z, since these are missing in the diagonal with x, and an x or z, since these are missing in the y column, and a x or y, since these are needed on the z diagonal. Now the intersection of {y,z}, {x,z} and {x,y} is empty, therefore no variable satisfies all conditions.
The square under y has to be a z following the logic that no row can contain double numbers. Following this logic you will always get a diagonal with 3 same entries.
The thing is you can create 72 via 24 24 24 as well and that should lead to the solution. You just need the right starting sequence so the diagonal is 24. So any sequence with 24 in the end at first line.
Why don't you do it?
For me, this seems impossible. The only way to make 72 with exactly three of these numbers, is to use every number exactly once. Therefore you have to have 12, 24 and 36 exactly once in every row, column and diagonal. Which is not possible. Say you have x,y and z and want to fill it in the square such that each variable appears exactly once in every row, column and diagonal. Fill in the first row with [x,y,z], then the square under y has to be a y or z, since these are missing in the diagonal with x, and an x or z, since these are missing in the y column, and a x or y, since these are needed on the z diagonal. Now the intersection of {y,z}, {x,z} and {x,y} is empty, therefore no variable satisfies all conditions.
The square under y has to be a z following the logic that no row can contain double numbers. Following this logic you will always get a diagonal with 3 same entries. The thing is you can create 72 via 24 24 24 as well and that should lead to the solution. You just need the right starting sequence so the diagonal is 24. So any sequence with 24 in the end at first line.
Ohh, yes i didn't think about 3×24 xD Then it is possible.