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 distance  2 is

A(1,1,1),B(2,3,5),C(−1,0,2) direction ratios of AB are <1,2,4⟩ Direction ratios of AC are <−2,−1,1>.  Therefore, direction ratios 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 be 2x−3y+z=k. Then k4+9+1=2  or  k=±214 Hence, equation of the required plane is 2x−3y+z+214=0