Consider $A,B,C,D,E,F,G $ are seven points in $R^3$ where no three points are collinear. Let $A,B,C $ be coplanar points lying in plane $P_1$ and $D,E,F,G $ be coplanar points lying in plane $P_2$ and out of the seven points no point lies on both the planes. Find the minimum number of pairs of skew lines joining any two of these points.
I started with cases. I got 12 pairs in the case in which 1 point is on one plane and other 3 are on the second plane. 36 in just the case opposite. The big problem is the 2 and 2 case. It's just very hard to figure out. I do think the minimum value will come when the 4 points form a trapezium and their non parallel sides intersect on the line which is the common line of both the planes. And then the three points of the triangle should be such that one the point is also on that common line. Just can't proceed further. You may think fresh.