If I got your problem correctly:
1. you have 4 points: A,B,C,D, not coplanar, and not even aligned by
group of 3. (A,B,C,D define a non-degenerated tetrahedron)
2. you want to find the sphere whose surface includes all the points.

You need additional constraint on D, or the sphere does not exist.

Notice that A,B,C & D are exchangeable. The sphere hold the ABC circle,
but also the ABD, ACD and BCD circles.

The normal going through the centre of all circles meets at the center
of the sphere.

So, one possible solution to check if there is a solution is:
 1. compute the circle for ABC: Center K, normal N.
 2. compute the circle for ABD (or any of the two others): Center L,
normal M.
 3. compute the smallest distance W between (K,K+N) line and (L,L+M) line
 4. if W is not null, there is no intersection, hence no solution.
 5. if W is null, the intersection I is the center of the sphere, and
the radius is IA, IB, IC or ID.


-- 
IQ of crossposters with FU: 100 / (number of groups)
IQ of crossposters without FU: 100 / (1 + number of groups)
IQ of multiposters: 100 / ( (number of groups) * (number of groups))

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