A plane P passes through the line of intersection of the planes 2...

To find, intersection point first put $z=0$\begin{aligned} & 2 x-y=2 \\ & x+y=1 \\ & \hline 3 x=3 \\ & x=1 \\ & x+y=1 \\ & y=1-x \end{aligned}At $x=1, y=0(x, y, z) \equiv(1,0,0)$ Putting $x=0\begin{aligned} & y-z=1 \\ & \frac{-y+3 z=2}{2 z=3} \\ & z=\frac{3}{2} \\ & y=1+z=1+\frac{3}{2}=\frac{5}{2} \\ & (x, y, z) \equiv\left(0, \frac{5}{2}, \frac{3}{2}\right) \end{aligned}$$$Point of intersection$\left(x_1, y_1, z_1\right) \equiv(1,0,0) \left( x_2, y_2, z_2\right)\equiv \left(0, \frac{5}{2},\frac{3}{2}\right)$Hence direction ratios of$the line of intersection of given plane\langle 2,-5,-3\rangle$.