TITLE: Number of planes that are equidistant from 4 non-coplanar points
QUESTION [6 upvotes]: If there are four non-coplanar points, find the number of planes such that all four of them are equidistant from the plane. Sorry one of those problems where 
I don't know what to do. How should I do this sum?

REPLY [6 votes]: There are two distinct scenarios. (This answer is not entirely rigorous, but it provides help with visualisation. Also, images are generated from GeoGebra, an amazing free and open source application. Note the specificity of the coordinates are irrelevant, picked for visual coherence.)
First scenario: choose any three points from the four, and form a plane through them. Then construct a perpendicular line from the fourth point to the plane, and find the midpoint of this line. The red plane through this midpoint, parallel to the blue plane, is equidistant to all 4 points. Note: There are $\binom {4}{3}$ occurrences of this scenario, since you are choosing three points from four.

Second scenario: Split the points into pairs. Then form parallel planes (blue) through each of the pairs. A third plane parallel to these two and in the center of them is equidistant to all 4 points. There are $1/2 \times \binom {4}{2} $ occurrences of this scenario, since you are choosing 2 pairs from four. Here's a link on choosing pairs: Combination of splitting elements into pairs

$$ \binom{4}{3} + \frac{1}{2} \binom{4}{2} = 4+3 = 7 $$