Sphere equation given 4 points Find the equation of the Sphere give the 4 points (3,2,1), (1,-2,-3), (2,1,3) and (-1,1,2).
The *failed* solution I tried is kinda straigh forward:
We need to find the center of the sphere.
Having the points:  
$$
p_{1}(3,2,1),\,
p_{2}(1,-2,-3),\,
p_{3}(2,1,3),\,
p_{4}(-1,1,2)
$$
2 Triangles can be created using these points, let's call $A$ our triangle formed by the points $p_{1},\,p_{2}\, and\,\, p_{3}$; And $B$ our triangle formed by the points $p_{1},\, p_{3}\, and \,\,p_{4}$.
Calculate the centroids of each triangle:
$$
CA = (2,1/3,1/3)\\
CB = (4/3,4/3,2)
$$
And also, a normal vector for each triangle:
$$
\overrightarrow{NA} = \overrightarrow{p_{1}p_{2}} \times \overrightarrow{p_{1}p_{3}}\\
\overrightarrow{NB} = \overrightarrow{p_{1}p_{3}} \times \overrightarrow{p_{1}p_{4}}
$$
$$
\overrightarrow{p_{1}p_{2}} = <-2,-4,-4>\\
\overrightarrow{p_{1}p_{3}} = <-1,-1,2>\\
\overrightarrow{p_{1}p_{4}} = <-4,-1,1>\\
\:\\
\overrightarrow{NA} = <-12, 8, -2>\\
\overrightarrow{NB} = <1, -7, -3>\\
$$
With the centroids and normals of triangles $A$ and $B$, we can build two parametric equations for a line, the first one intersects the centroid of $A$ and the other one the centroid $B$.
$$
Line \enspace A\\
x = 2-12t \quad
y = 1/3+8t \quad
z = 1/3-2t\\
\:\\
Line \enspace B\\
x = 4/3 + s \quad
y = 4/3 - 7s \quad
z = 2 - 3s
$$
The point where this lines intersect should be the center of the sphere, unfortunately this system of equations is not linearly dependent, that means that they do not intersect each other. What could be the problem here?

I would cite the beautiful method from W.H.Beyer to find the center and radius of the sphere $(x-a)^2+(y-b^2)+(y-c)^2=R^2$

