The plane `x-2y+3z=0` is rotated through a right angle about the line of intersection with the plane `2x+3y-4z=0,` find the equation of the plane in its new position.

Explanation

Let the equation of the plane be
`x−2y+3z+λ(2x+3y−4z−5)=0`

`⇒(1+2λ)x+(−2+3λ)y+(3−4λ)−5λ=0 ...(1)`

If the plane is perpendicular to the plane `x−2y+3z=0`, then
`1.(1+2λ)−2(3λ−2)+3(3−4λ)=0`

`⇒λ=7/8`

Thus, the required equation is
`22x+5y−4z=35`