Skrytý text:first thing to realize is that for given 4 points none of them will belong to the plane we look for;
if there was at least one such point, its distance from the plane is zero
and the need for equidistance will put all 4 points into this plane which makes them coplanar
so all the points lie outside the plane that we look for
another thing is that such plane would split the 3D space into two half-spaces
and each half-space contains at least one of the given 4 points, if they were all in one half-space,
that would necessarily make them coplanar because of equidistancy
so we can have two points in each half-space or one point in first one and 3 points in other one
let's see the case where points AB are in first half-space and CD in the other one
we can construct lines
and
those two lines does not intersect as in such case the 4 points would be coplanar
the plane we look for has the normal
giving the equation
where only
is not known
now there exists
for which the line
lies in the plane
and also
such that line q lies in the plane
planes
are parallel since they share normal vector, let be
there is
,
also
now, moving
between constants
gives us equations for planes
parallel to
with
where the left side grows from
to
and the right side decreases from
to
there is one plane there for which we get equality of the distances in the equation which gives us one and only one plane
as there are for ways how to distribute 4 points by two in each half-space, this gives us 4 plains
now let be
the point in first half-space and
points in the second half-space
those 3 points form the plane
and for sure there is unique plane
going through
parallel to
again there is one unique plane
between those two where
as there are 4 ways to put only one of given point into the first half-space, this gives us 4 more planes
so the resulting number of planes should be 7
the assumption that no 3 of points are colinear is not necessary to add, collinearity of 3 points would make
the four points coplanar