Minimal planes defined by n non-collinear 3D points? Is there a set of 3D points where every plane of 3 points also has a 4th point?
According to La Jolla Covering Repository, the following sets of $n$ points and $k$ 4-planes might be possible: 
$$(8, 14), (10, 30), (14, 91), (16, 140), (20, 285), (26, 650), (28, 819)$$
I'll add two more condition:  no three points can be collinear, and there must be at least 4 planes defined by the points.
With vertices of a cube there are 8 points with 12 planes of 4 points, but there are also 8 planes with 3 points.  
For $n$ points and $k$ 5-point planes, the following might be possible:  
$$(17, 68), (26, 260)$$
For $n$ points and $k$ 6-point planes, the following might be possible:
$$(22, 77), (26, 130)$$
The $(22,77)$ case would equivalent to the 77-graph on  Mathieu group $M_{22}$.
abcilu  abdfrs  abejop  abgmnq  abhktv  acdghp  aceqrv   
acfjnt  ackmos  ademtu  adinov  adjklq  aefgik  aehlns
afhoqu  aflmpv  agjsuv  aglort  ahijmr  aipqst  aknpru
bcdekn  bcfgov  bchjqs  bcmprt  bdgijt  bdhlmo  bdpquv
beflqt  beghru  beimsv  bfhinp  bfjkmu  bgklps  bikoqr
bjlnrv  bnostu  cdfimq  cdjoru  cdlstv  cefpsu  cegjlm
cehiot  cfhklr  cginrs  cgkqtu  chmnuv  cijkpv  clnopq
defhjv  degoqs  deilpr  dfglnu  dfkopt  dgkmrv  dhiksu
dhnqrt  djmnps  efmnor  egnptv  ehkmpq  eijnqu  ejkrst
eklouv  fghmst  fgjpqr  fijlos  firtuv  fknqsv  ghilqv
ghjkno  gimopu  hjlptu  hoprsv  iklmnt  jmoqtv  lmqrs

Or perhaps discard the condition that each plane must have the same number of points.  Is there a 3D set of points with no 3-planes?
Or perhaps just fewest planes with $n$ points.
4 points:  tetrahedron, 4 planes.
5 points: 4-pyramid, 7 planes.
6 points: octahedron, 3-prism, 5-pyramid; 11 planes.
7 points: 6-pyramid, 16 planes?
8 points: cube, 20 planes?  
Or maybe drop the non-collinearity requirement.  Is there a set of points in 3D where all defined planes have more than 3 points?
 A: The answer is no; every set of points in three space, with no three collinear and not all contained in a plane, has a plane which intersects exactly $3$ of the points. To prove this, consider the point $P$ with the lowest $z$ coordinate. Say that $P$ has a negative $z$-coordinate, and all other points have a positive $z$-coordinate. Now, for each other point $Q$, let $Q'$ be the intersection of $\overline{PQ}$ with the $xy$-plane. By the Sylvester-Gallai theorem, there is a line $\ell$ in the $xy$-plane which intersects exactly $2$ of these of points, $Q'$ and $R'$. This means the plane containing $\ell$ and $P$ intersects exactly $3$ of the original points, $P,Q$ and $R$.
When you drop the collinearity requirement, it is possible. Consider two skew lines, with three points on each.
I do not know about what the fewest number of planes defined by $n$ points is. However, this article has some information about the minimal number of $3$-planes defined by $n$ points:
https://arxiv.org/abs/1608.03189
