If the line $y-x+2=0$is shifted parallel to itself towards the positive direction of the  x-axis by a perpendicular distance $3\sqrt{2}$ units, then the equation of the new line is
 

Given line y-x+2=0
$\mathrm{x}-\mathrm{y}-2=-0....\left(1\right)$
$\mathrm{x}-\mathrm{y}+\mathrm{\lambda }=-0....\left(2\right)$    be required line

Given thatdistance between (1) and (2) is $3\sqrt{2}$
$$\begin{gathered} \Rightarrow \left|\frac{\mathrm{\lambda }+2}{\sqrt{2}}\right|=3\sqrt{2} \\ \Rightarrow \mathrm{\lambda }=-8,4 \end{gathered}$$
Line shifted to +ve  x axis so line  is  x-y-8=0   or y=x-8