If the plane $x+y+z=1$ is rotated through an angle ${90}^0$ about its line of intersection with the plane  $x-2y+3z=0$, then the new position of the plane is 

$x+y+z=1\left(1\right)$

Required plane $x+y+z-1+\alpha \left(x-2y+3z\right)=0$

$\left(1+\alpha \right)x+\left(1-2\alpha \right)y+\left(1+3\alpha \right)z-1=0\left(2\right)$

(1) and (2) are perpendicular

$\Rightarrow \left(1\right)\left(1+\alpha \right)+\left(1\right)\left(1-2\alpha \right)+\left(1\right)\left(1+3\alpha \right)=0\Rightarrow 2\alpha =-3\Rightarrow \alpha =\frac{-3}{2}$

$\therefore$Required plane is 

$x+y+z-1\frac{-3}{2}\left(x-2y+3z\right)=0$

$2x+2y+2z-2-3x+6y-9z=0$

$-x+8y-7z=2,x-8y+7z=-2$