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anilorap
suppose P=(2,-4,3) Q=(0,-1,2) and R=(1,1,1). Find the plane PQR
First, find two vectors in the plane PQR: PQ=<0-2,-1-(-4),2-3>=<-2,3,-1> PR=<1-2,1-(-4),1-3>=<-1,5,-2> PQxPR is the normal to the plane PQR: i j k -2 3 -1 -1 5 -2 =<-1,-3,-7> The plane is defined by the normal passing through one of the three points, we can choose (1,1,1) for convenience: -1(x-1) -3(y-1) -7(z-1)=0 which simplifies to x+3y+7z-11=0 as the equation of the required plane. Check that all points PQR lie in the plane.