Angles made with the $x-$axis by two lines drawn through the point $(1, 2)$ and cutting the line $x + y = 4$ at a distance $\frac{\sqrt{6}}{3}$ from the point $(1, 2)$ are 

Suppose a line passing through $(1, 2)$ makes an angle $\theta$ with $x-$axis. Then, its equation is

          $\frac{x-1}{\cos \theta }=\frac{y-2}{\sin \theta }$

The coordinates of a point on this line at a distance $\frac{\sqrt{6}}{3}=\sqrt{\frac{2}{3}}$ from $(1, 2)$ are given by

 $\frac{x-1}{\cos \theta }=\frac{y-2}{\sin \theta }=\sqrt{\frac{2}{3}}$ or, $\left(1+\sqrt{\frac{2}{3}}\cos \theta ,2+\sqrt{\frac{2}{3}}\sin \theta \right)$

This point lies on $x+y=4.$

$\therefore 1+\sqrt{\frac{2}{3}}\cos \theta +2+\sqrt{\frac{2}{3}}\sin \theta =4$

$\Rightarrow \cos \theta +\sin \theta =\sqrt{\frac{3}{2}}$

$\Rightarrow \sin \left(\theta +\frac{\pi }{4}\right)=\frac{\sqrt{3}}{2}\Rightarrow \theta +\frac{\pi }{4}=\frac{\pi }{3},\frac{2\pi }{3}\Rightarrow \theta =\frac{\pi }{12}$ and $\frac{5\pi }{12}$