Let the plane passing through point (2,1,-1) containing line joining the points (1,3,2) and (1,2,1) makes intercepts p,q,r on co-ordinate axes, then p + q + r =

Step-by-step solution

Three points on the plane are given: A = (2,1,-1) , B = (1,3,2) , and C = (1,2,1) . Vectors in the plane are AB = (-1,2,3) and AC = (-1,1,2) . The normal vector n is perpendicular to AB and AC , found using the cross product: n = AB AC = i (2 2 - 3 1) - j (-1 2 - 3 (-1)) + k (-1 1 - 2 (-1)) = (1, -1, 1) . Using point A and normal n = (1,-1,1) , the plane equation is 1(x-2) -1(y-1) +1(z+1)=0 . Simplifying gives x - y + z = 0 . The intercepts on the axes are all zero since setting any two variables to zero forces the