Let  A(1,1,1),B(2,3,5) and C(−1,0,2) be three points, then equation of a plane parallel to the plane ABC which is at a distance  2 units is

A(1,1,1),B(2,3,5),C(−1,0,2) d.r’s of AB  are (1, 2, 4) D.r’s of AC are (−2,−1,1). D.r’s of normal to plane ABC are (2,-3,1) As a result, equation of the plane ABC is 2x−3y+z=0. Let the equation of the required plane is , 2x−3y+z=k then k4+9+1=2k=±214Hence, equation of the required plane is 2x−3y+z+214=0