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  $\left|\frac{\mathrm{k}}{\sqrt{4+9+1}}\right|=2\text{ or }\mathrm{k}=\pm 2\sqrt{14}$

Hence, equation of the required plane is  $2\mathrm{x}-3\mathrm{y}+\mathrm{z}+2\sqrt{14}=0$