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, l >. 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=±214Hence, equation of the required plane is  2x−3y+z+214=0