Planes

Let the points be $A,B,C \ and \ D$ .

No. of planes which have 3 points on one side and the fourth point on the other side=4

No. of planes which have two points on each side of plane =3

Hence, total no. of the required planes=4+3= $\boxed{7}$

The solution by the author is very good, but I felt that it doesn't explain much and leaves a lot of room for doubt. So, here goes:

Observe that for points to be equidistant from a plane, all of them have to lie on either of the two planes which are parallel to the said plane.

Now, we can see that if we take any $3$ points, then we can always find a plane $P2$ parallel to the plane of these points such that the distance between the plane of these points and $P2$ is same as distance between $P2$ and the fourth point. So this configuration gives us $4$ planes, which is the number of distinct ways of selecting $3$ points from the given $4$ .

It is known that, when there are two vectors in $3-D$ space, one can always find a plane $P$ such that this plane is parallel to both vectors(using cross product). Now, if we are to divide the $4$ given points into two pairs and join(exclusively) the points of a pair, we end up with two non co-planar vectors. We can always find a plane which is parallel to these vectors. If we shift the said plane parallel along its normal, it's obvious that in a certain configuration, the plane will be equidistant from all the four points. So, this configuration gives us $3$ more planes, which is the number of ways of selecting two distinct pairs from $4$ .

Thus, the answer is $4+3=\boxed{7}$

Let me try my hand in offering an intuitive "proof". First, given that $4$ non-colinear points forms an irregular tetrahedron, it's easy to see that no such plane can exist outside it. Next, any $3$ points forms a plane which is parallel to a plane that is halfway to the $4$ th point, so we have $4$ such planes. Now, the tricky step. There are $3$ pairs of points of the tetrahedron between which there could exist such a plane. To prove that there will always be one such plane, imagine that the lines through each pair of points are replaced with cylinders of the same radius, and we let the radius expand until the cylinders touch. There's your plane, tangent to both cylinders where they touch, and so we have $3+4$ or $7$ planes in all.

Finding a plane that is equidistant from the four points A, B, C, D is equivalent to finding a pair of parallel planes such that all of the points lie on one of the planes. Suppose we find such a pair of planes. One of them must contain A. Call it 1. Call the other one as 0. B, C and D must each lie on either of these planes, but the one where all lie on 1 is ruled out as the points are non-coplanar.

Hence, number of planes is 2^3 - 1 = 7.

There are 4 planes that contains three points. And there are 3 planes that separate these four points in two pairs.

$\frac{_4 C_3}{1} + \frac{_4 C_2}{2}$ = 4 + $\frac62$ = 7

There are 0 for 4 + 0;

There are 4 for 3 + 1;

There are 3 for 2 + 2;

For 2 + 2, they shall always form skewed lines in 3 dimensional space which are equivalent to two parallel lines being twisted with an angle which is not zero onto its "H" shape. Certain views from proper orientations allow us to see this.

Answer: $\boxed{7}$

Found the same question in 'Problem solving strategies'

Imagine a cube for ease, place points on daigonal ends of opposite face to get 4 points non coplanar, center of cube as (0,0,0), there are 3 planes which has 2 points on one side , make tetrahedron by choosing any 3 as base given by 4C3 to get 4 tetrahedron => 4+3=7.