Two lines $L_1$and $L_2$ passing through the point $P(1,2)$ cut the line $x+y=4$ at a distance of $\frac{\sqrt{6}}{3}$ units from P. Then the angles made by$L_1,L_2$with positive X-axis are

Parametric point is

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

The point lies on $x+y=4$

$3+\sqrt{\frac{2}{3}}(\sin \theta +\cos \theta )=4$

Squaring on both sides will get,

$\begin{array}{r}{\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =\frac{3}{2}\\ 1+\sin 2\theta =\frac{3}{2}\\ \sin 2\theta =\frac{1}{2}\end{array}$

$=\frac{\pi }{12},\frac{5\pi }{12}$