1. The problem statement, all variables and given/known data Question states "The plane that contains the line r=<-2,4,3>+t<3,2-1> and is perpendicular to the plane r=<5,0,0>+s<2,1,0>+t<-1,0,1> is:" Answer is y+2z=10 2. Relevant equations Cross product and dot product of vectors 3. The attempt at a solution I found a vector normal to the plane r=<5,0,0>=s<2,1,0>+t<-1,0,1> by doing the cross product of the two direction vectors is <2,1,0> x <-1,0,1> getting <1,-2,1> than apply rule of n.v=0 ie <1,-2,1> . <x-(-2), y-4, z-3> to get x+2-2y+8+z-3=0 and so x-2y+z = -7 Not sure what I have done wrong here, could someone explain please?