Imagine you had a sphere of radius R centered at the origin. What are the coordinates of the vertices of the regular tetrahedron which is circumscribed by the sphere? One of the vertices of the tetrahedron is (0,0,R) and one of the vertices lies in the z,x plane.
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A hint rather than a proper answer: alternating vertices of a cube (e.g. the vertices $(-1, -1, -1), (-1, 1, 1), (1, -1, 1),$ and $(1, 1, -1)$ of the cube $[-1..1]^3$) form the vertices of a regular tetrahedron. This allows you to easily calculate the internal angle of the tetrahedron (i.e., the angle between the lines from the center to any two vertices), and that internal angle provides the position of the vertex in the $xz$ plane. Once you have that vertex, you can find the others by rotating its position $\pm120$ degrees about the $z$ axis.
I found at this site
The tetrahedron coordinates:
Vertex coordinate 0, x= 0.000, y= 0.000, z= 1.000 1, x= 0.943, y= 0.000, z=-0.333 2, x=-0.471, y= 0.816, z=-0.333 3, x=-0.471, y=-0.816, z=-0.333
Length of every edge 1.6329932
Of course, the real answers have square roots in them. You can scale by the radius