Is it possible to have a triangle in 3D space where:
[1] Side lengths are all integers,
[2] Coordinates of vertices are all integers, and
[3] No two vertices share an x,y, or z coordinate?
Counterexample: The triangle with vertices at (0,0,0), (0,5,0) and (4,0,0) satisfies [1] and [3] above because it is a 3/4/5 right triangle with integer coordinates, but fails [3] because two vertices share an x coordinate of zero (two share a y of zero and all share a z of zero, as well).
The face diagonals of an Euler brick would work, if it's possible to rotate the brick in space such that at least three of its corners land on integer coordinates.
Edited to add: in a comment on a since-deleted incorrect answer, a user (forgot your name, sorry!) pointed out that for a Pythagorean triplet [e,f,g], the following points satisfy the question conditions:
(0,0,0) (e,2e,2e) (2f,f,-2f) for side lengths 3e, 3f, 3g
So yes, it is possible. There are additional valid formulas:
(0,0,0) (3e,4e,12e) (12f,3f,-4f) for side lengths 13e, 13f, 13g
(0,0,0) (4e,5e,20e) (20f,4f,-5f) for side lengths 21e, 21f, 21g
(0,0,0) (2e,3e,6e) (6f,2f,-3f) for side lengths 7e, 7f, 7g