If such a pyramid exists, could someone provide the coordinates of its vertices?

$\begingroup$ Here’s a related Concept : en.m.wikipedia.org/wiki/Trirectangular_tetrahedron $\endgroup$– Sidharth GhoshalSep 26 at 16:00

$\begingroup$ See also math.stackexchange.com/questions/3214139/… $\endgroup$– Marco RipàSep 26 at 16:06

4$\begingroup$ "All four faces" seems to imply you mean a tetrahedron. If you had said "tetrahedron" rather than "pyramid", this would be clearer. $\endgroup$– Michael HardySep 26 at 23:33

$\begingroup$ If three right angles are at one vertex, then you just have the corner of one octant of the usual $(x,y,z)$Cartesian coordinate system. Are there points $(x>0,0,0),\,\, (0,y>0,0),\,\, (0,0,z>0)$ that are the vertices of a right triangle? No. Now the alternatives are: two right angles and one acute angle at one vertex, or at most one right angle at each vertex. $\endgroup$– Michael HardySep 26 at 23:38

2$\begingroup$ en.wikipedia.org/wiki/Schl%C3%A4fli_orthoscheme $\endgroup$– Robin HoustonSep 27 at 0:34
3 Answers
If by pyramid, you mean the shape with a square base and four triangular sides, and you want the right angles all where the triangles meet, there is a solution, but it is degenerate.
Sample Vertices: Set A: (0,0,0), (1,0,0), (0,1,0), (1,0,0), (0,1,0) Set B: (0,0,0), (1,1,0), (1,1,0), (1,1,0), (1,1,0)
By degenerate, I mean that all of the vertices are in the same plane (z=0, in my example), and the shape has zero volume. It's flat.
It is obvious that this must be the case. You have four right angles meeting at a point. That's 360 degrees, exactly what you need for them to be in the same plane.
If the first point in my example A was (0,0,1), then all of the edges would be the same length (the square root of 2), so the triangles would all be equilateraland all of their angles would be 60 degrees. The more you raise that point, the smaller the angles there get.
Giving this StackExchange one last try. In addition to the answer posted with triangular base, here is an example with a square base:
Use two 345 right triangles and two 35sqrt(34) triangles.