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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.

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At most 2 of the pairs of lines (DE,FG),(DF,EG),(DG,EF) intersect at the intersection of plains $P_1,P_2$. Let say the first 2 of the 3, and they intersect at points M and N. Let M$\in$AB and N$\in$AC. So the fours ABDE, ABFG, ACDF, ACEG are coplanar, and we can pick BC so BCDG are coplanar

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I miscounted some pairs( a lot of them...), so in general case there can be 1) one line in each plain P1 and P2, $3 \choose 2$$4 \choose 2$=18, 2) one line connects 2 points from P1 and P2 each and other is in one of the plains $3 \choose 1$$4 \choose 1$($2 \choose 2$+$3 \choose 2$)=48 3) both lines connect different points from P1 and P2 2$3 \choose 2$$4 \choose 2$=36

So maximum umber of pairs is 18+48+36=102, minus 5 times 3 pairs we dismissed is 87

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