Three planes meet at two points it means they have infinitely many solutions, so  $\left|\begin{array}{ccc}2& 1& 1\\ 1& -1& 1\\ \alpha & -1& 3\end{array}\right|=0$ 

$\Rightarrow 2(-3+1)-1(3+1)+\alpha (1+1)=0\Rightarrow \alpha =4$

$P_1:2x+y+z=1;P_2:x-y+z=2;P_3:4x-y+3z=5$

$P\text{ on }XOY\text{ plane }=(1,-1,0)$

(which can be obtained by putting z=0 in any two of the given planes)
Q on YOZ plane $=\left(0,\frac{-1}{2},\frac{3}{2}\right)$

(which can be obtained by putting x=0 in any two of the given planes)

$\therefore$ straight line perpendicular to plane $P_{3 }$passing through P is:

$\frac{x-1}{4}=\frac{y+1}{-1}=\frac{z}{3}$

$\overrightarrow{PQ}=\widehat{i}-\frac{1}{2}\widehat{i}-\frac{3}{2}\widehat{k}$

Projection of $\overrightarrow{PQ}$ on x-axis $\Rightarrow \left|\frac{\overrightarrow{OP}.\widehat{i}}{\left|\widehat{i}\right|}\right|=1$

Centroid of $\mathrm{△}OPQ$ is $\left(\frac{1}{3},\frac{-1}{2},\frac{1}{2}\right)$