If the plane x + y + z = 1 is rotated though an angle 90° about its line of intersection with the plane x – 2y + 3z = 0, then the new position of the plane is

 $$\begin{gathered} Let equation plane be (\mathrm{x}-2\mathrm{y}+3\mathrm{z})+\mathrm{\lambda }(\mathrm{x}+\mathrm{y}+\mathrm{z}-1)=0 \\ It is perpendicular to the plane x+y+z=1 \end{gathered}$$

$$\begin{gathered} \begin{array}{lc} & \therefore (1+\mathrm{\lambda })+(\mathrm{\lambda }-2)+(\mathrm{\lambda }+3)1=0 \\ \Rightarrow 3\mathrm{\lambda }+2=0\Rightarrow \mathrm{\lambda }=\frac{-2}{3} \\ (\mathrm{x}-2\mathrm{y}+3\mathrm{z})-\frac{2}{3}(\mathrm{x}+\mathrm{y}+\mathrm{z}-1)=0\\ & 3\mathrm{x}-6\mathrm{y}+9\mathrm{z}-2\mathrm{x}-2\mathrm{y}-2\mathrm{z}+2=0 \\ \Rightarrow \mathrm{x}-8\mathrm{y}+7\mathrm{z}=-2\end{array} \end{gathered}$$