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 a distance  2 units is

$A(1,1,1),B(2,3,5),C(-1,0,2)$ d.r’s of AB  are (1, 2, 4) 
D.r’s of AC are $(-2,-1,1)$. 
D.r’s 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 is , $2x-3y+z=k$ then $\left|\frac{{\displaystyle k}}{{\displaystyle \sqrt{4+9+1}}}\right|=2k=\pm 2\sqrt{14}$

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