Slope of a line $L$  is 1. The points on the line $L$ at a distance of 2 units from the point $\left(3\sqrt{2},3\sqrt{2}\right)$ which lies on the line $L$ are

$\text{ The inclination of the line }L\text{ is }\theta ={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$

$\text{ The coordinates of a point on the line }L\text{ at a distance of }r\text{ units from }\left(x_1,y_1\right)\text{ are }$

$x=x_1\pm r\cos \theta \text{ and }y=y_1\pm r\sin \theta ,\text{ here }\theta \text{ is inclination of the line }\theta$

$x=x_1\pm r\cos \theta \text{ and }y=y_1\pm r\sin \theta ,\text{ here }\theta \text{ is inclination of the line }\theta$

Hence the required coordinates are 

$\begin{array}{c}x=x_1\pm r\cos \theta \\ =3\sqrt{2}\pm 2\left(\frac{1}{\sqrt{2}}\right)\\ =3\sqrt{2}\pm \sqrt{2}\\ =4\sqrt{2},2\sqrt{2}\end{array}$

and 

$\begin{array}{c}y=y_1\pm r\sin \theta \\ =3\sqrt{2}\pm 2\left(\frac{1}{\sqrt{2}}\right)\\ =3\sqrt{2}\pm \sqrt{2}\\ =4\sqrt{2},2\sqrt{2}\end{array}$

$\text{ Therefore, the required points are }(4\sqrt{2},4\sqrt{2})=(2\sqrt{2},2\sqrt{2})$