For a tetrahedron, the circumsphere is calculated, then it's recentered at the origin and the tangent planes at the vertices are found. Plane intersections are found, then the recentering is reversed.
initial = {{0, 0, 0}, {1, 2, 0}, {3, 0, 0}, {1, 1, 2}};
cs = Circumsphere[initial];
eq = Append[(# - cs[[1]]), -cs[[2]]] . {x, y, z, cs[[2]]} & /@ initial;
tangential = # + cs[[1]] & /@ Flatten[{x, y, z} /. Solve[#[[1]] == #[[2]] == #[[3]] == 0] & /@ Subsets[eq, {3}], 1];
tan = {{3/2, 1/2, -5}, {-(17/18), 31/18, 10/9}, {3/2, -(21/2), 6}, {31/6, 25/6, 7/3}};
That finds the tangential tetrahedron, which is tangent to the circumsphere of the given tetrahedron and which has an insphere the same as the circumsphere.
This is the right solution, but my method feels like a B- level hack. Is there a more elegant method, with pure linear algebra or trilinear coordinate methods?