Find the equation of the plane through the line of intersection of the planes `x+y+z=1` and `2x+3y+4z=5`, which is perpendicular to the plane `x-y+z=0`. Also find the distance of the plane so obtained from the origin.

Explanation

Let the required equation of the plane be
`(x+y+z-1)+lambda(2x+3y+4z-5)=0`
For some real value of `lambda`.
`rArr (1+2lambda)x+(1+3lambda)+(1+3lambda) + (1+ 4lambda)z-(1+5lambda)=0`………………..(i)
Since, this plane is perpendicular to the plane `x-y+z=0`, we have
`(1+2lambda) xx 1 + (1 + 3lambda) xx (-1) +(1+4lambda) xx 1=0`.
`rArr (1+2lambda)-1/3z+2/z=0 rArr x-z+2=0`.
Hence, the required equation of the plane is `x-z+2=0`.
Distance of the plane from the origin.
= Length of perpendicular from the origin to the plane
`=|0-0+2|/sqrt(1^(2)+0^(2)+(-1)^(2))=2/sqrt(2)=sqrt(2)`.