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

On solving $mx-y+(2-m)=0$
and $x+y=4,$  we get
$(x,y)=\left(\frac{{\displaystyle m+2}}{{\displaystyle m+1}}\frac{{\displaystyle 3m+2}}{{\displaystyle m+1}}\right)$
Now, it is given that
 $\begin{array}{l}\sqrt{{(\frac{m+2}{m+1}-1)}^2+{(\frac{3m+2}{m+1}-2)}^2}=\frac{\sqrt{6}}{3}\\ \Rightarrow \frac{1}{{(m+1)}^2}+\frac{m^2}{{(m+1)}^2}=\frac{6}{9}=\frac{1+m^2}{{(m+1)}^2}=\frac{2}{3}\end{array}$
$$\begin{gathered} \Rightarrow 3+3m^2=2m^2+4m+2 \\ \begin{array}{l}\Rightarrow m^2-4m+1=0\\ \Rightarrow m=2\pm \sqrt{3}\Rightarrow m=2+\sqrt{3},2-\sqrt{3}\end{array} \\ \therefore Anglewillbe\frac{\pi }{12}and\frac{5\pi }{12} \end{gathered}$$