$\text{ The inclination of the line joining the centroid of the triangle formed by the points }$$(1,2),(2,1),(3,6)\text{ and the point of intersection of lines }x+y-2=0\text{ and }x-y=0\text{ is }$

$\text{ Centroid of triangle formed by }A\left(x_1,y_1\right),B\left(x_2,y_2\right),C\left(x_3,y_3\right)\text{ is }G\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

$\text{ Hence the centroid of the triangle formed by the points }(1,2),(2,1),(3,6)\text{ is }G=(2,3)$

To get the point of intersection of two lines, eliminate   by adding two equations given.  

$\begin{array}{c}x+y-2+x-y=0\\ 2x=2\\ x=1\end{array}$

$\text{ Substitute }x=1\text{ in }x-y=0$

$\text{ it implies, }1-y=0\Rightarrow y=1$

$\text{ Hence the point of intersection of two lines is }(1,1)$

$\text{ The slope of the line joining two points }(1,1)\text{ and }(2,3)\text{ is }m$

$\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{3-1}{2-1}\\ =\frac{2}{1}\\ =2\end{array}$

$\text{ If the inclination of any line }\theta \text{ then the slope of that line is }m=\tan \theta$

$\text{ Here, }\tan \theta =2\Rightarrow \theta ={\tan }^{-1}\left(2\right)$

$\text{ Therefore, the inclination is }{\tan }^{-1}\left(2\right)$