This is the easy way if you can’t do simple vector logic:

Keep in mind that the first vector must have all the minimal values you want to check and second all the largest values of a vector. ( Or just see second example )

So i have to invert the two previous vector ? Like that ?
print(PlayerPos:WithinAABox(Vector(7663,-7663, 96),Vector(7078, -7154, 0)))
because it still print me false

Yeah, I got confused with WithinAABox as well. The first vector argument needs to be smaller IN ALL OF ITS VALUES than the second one. So x1 < x2 y1 < y2 z1 < z2 etc.