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

$A(1,1,1),B(2,3,5),C(-1,0,2)\text{ direction ratios of }AB\text{ are }<1,2,4⟩$

$\text{ Direction ratios of }AC\text{ are }<-2,-1,1>\text{. }$

$\text{ Therefore, direction ratios of normal to plane }ABC\text{ are }<2,-3\text{, 1>}$

$\text{ As a result, equation of the plane }ABC\text{ is }2x-3y+z=0\text{. }$

$\text{ Let the equation of the required plane be }2x-3y+z=k\text{. Then }$

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

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