A Minimum Path to visit $4$ points on a sphere (no intersections)

I have a question:

There is a tetrahedron with coordinates $(3,1,5),(-7,0,13),(27,5,-13),(-17,4,21)$ that is inscribed in one sphere. What is the minimum distance needed to "walk" to each point once without a diagonal intersection (to the nearest hundredth since I don't want messy answers)?

I created this question to test my geometry.

I figured I could use the Law of Cosines with Extended Law of Sines to find the circumcircle of any set of three points, but I can't figure out where to go from those circles.

Right now, I need to find where the sphere is, and how to find the minimum perimeter from there.

Note: There is actually one. Oops.

Update: As Narlin has shown, the sphere's center is $(393,\frac{69}2,2507),$ with an radius of approximately $636.57383\dots.$ The task is now down to finding the minimum perimeter.

I translating the sphere and tetrahedron such that the origin is the center of the sphere is useful. Now it's just a matter of finding the shortest path.

All we care about is the angle measurements between points (and also intersections).

I can use the vector formula $\textbf{v}\bullet\textbf{w}=||\textbf{w}||||\textbf{v}||\cos{\theta},$ but this, along with checking combinations, seems a bit too long to me. Is there a way to use a shortest distance result, or is my method the best?

• Do you mean the sphere passing through those points? But what quadrilateral are you talking about? Your question is not clear: please improve it. – Aretino Sep 13 '18 at 11:55
• Okay... I edited it... – Jason Kim Sep 13 '18 at 14:00
• Four non-coplanar points determine a unique sphere: why do you say there are two? – Aretino Sep 13 '18 at 20:54
• Oh... I thought that it could make 2.... I thought that there are two possible "rings" for 3 points + 1 other point that can be one of two rings... – Jason Kim Sep 13 '18 at 22:56