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
Hence, equation of the required plane is