A cube has 8 non-coplanar points at the vertices and has 6 faces. This is only a partial answer...3 points determine a plane so there will be many more than 6. Your answer is going to be found by the formula n!/(n-r)! where n=8 and r=3.
That gives: 40320/120 = 336
3
10!
one
Just one plane.
Exactly one plane in each case.
3
10!
one
Just one plane.
3 non-collinear points define one plane.
If points A, B, and C are not on the same line, they determine a single plane.
exactly one and only one.
Exactly one plane in each case.
1 line cause every plane contains atleast 3 or more noncollinear points
The answer depends on the number of point. One point - as the question states - cannot be non-collinear. Any two points are always collinear. But three or more points will define a plane. If four points are non-coplanar, they will define four planes (as in a tetrahedron).
If 2 points determine a line, then a line contains infinitely many planes.
Infinitely many planes may contain the same three collinear points if the planes all intersect at the same line.