If we substitute a length-7 stick for the length-1 stick, there are five unique ways to make the tetrahedron. The constraints are that the lengths on a side with the length-2 stick must differ by only 1, and (3,4) cannot be on the same side with the 7.
[Assume you're looking down/up at the figure. These solutions assume rotations and reflections are the same:]
/ \ / | \ 2 3 5 / | \ / / \ \ / 4 7 \ /_/____6____\_\ |
/ \ / | \ 2 3 6 / | \ / / \ \ / 4 7 \ /_/____5____\_\ |
/ \ / | \ 2 3 6 / | \ / / \ \ / 4 5 \ /_/____7____\_\ |
/ \ / | \ 2 3 7 / | \ / / \ \ / 4 5 \ /_/____6____\_\ |
/ \ / | \ 2 4 6 / | \ / / \ \ / 5 3 \ /_/____7____\_\ |
Each side is part of two triangles. Starting with the largest integer, one can find ONLY two triangles that can coexist: 6,5,2 and 6,4,3. This leaves the 1, which is not long enough to make two triangles, either 1,5,4 and 1,2,3, or 1,5,2, and 1,3,4. Therefore the original integers can not be used.
Changing the offender 1 to a 7 (the smallest integer greater than 6) would give us more options. Taking the other five integers and putting the largest with the smallest of them, we get 7,6,2 with the 3,4,and 5 left over. The 3 & 4 can't make a triangle with the 7, but the 5 and 3 can. (So can the 5 & 4. No difference.) This leaves the 4 with the triangles 7,6,2, and 7,5,3. You can make the 4 fit in EITHER the set of 4,5,6 and 4,3,2 OR the set of 4,6,3 and 4,5,2, giving you the desired solution.