Equation of a sphere as the determinant of its variables and sampled points

The equation of a sphere of radius $r$ centred at $(u,v,w)$ is $(x - u)^2 + (y-v)^2 + (z-w)^2 = r^2$, or $x^2 + y^2 + z^2 + U x + V y + W z + C = 0$ where $C = u^2 + v^2 + w^2 - r^2$, $U = -2u$, $V=-2v$, $W=-2w$. If $A$ is the matrix you're taking the determinant of, to have this equation satisfied for the four given points and an arbitrary point $(x,y,z)$ on the sphere means $$A \pmatrix{1\cr U\cr V\cr W\cr C\cr} = \pmatrix{0\cr 0\cr 0\cr 0\cr0\cr}$$ But for that to happen, $\det A$ must be $0$. Conversely, if $\det A = 0$, there exist $T,U,V,W,C$ not all $0$ for which $$A \pmatrix{T\cr U\cr V\cr W\cr C\cr} = \pmatrix{0\cr 0\cr 0\cr 0\cr0\cr}$$
If $T \ne 0$ we may divide by it and get an equation of a sphere going through the points. If $T = 0$ we have a degenerate case where the "sphere" is actually a plane.

Yes, this generalizes to any number of dimensions.