Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati+3hatj)-6=0` and `vecr.(3hati-hatj-4hatk)=0`, whose perpendicular distance from the origin is unity.

Explanation

Taking `vecr=(xhati+yhatj+zhatk)`, the equations of the given planes are
`(xhati+yhatj+zhatk). (hati+3hatj)-6=0 rArr x+3y-6=0`……….. (i)
(xhati+yhatj+zhatk).(3hati-hatj-4hatk)=0 rArr 3x-y-4z=0`………(iii)
Length of the perpendicular from the origin to plane (iii) is given as 1.
`therefore |0+0-0-6|/sqrt((1+3lambda)^(2)+(3-lambda)^(2)+(-4lambda)^(2))=1`
`rArr (1+3lambda)^(2)+(3-lambda)^(2)+)-4lambda)^(2)=36`
`rArr 26lambda^(2)=26 rArr lambda^(2)=1 rArr lambda=+1`.
Putting, `lambda=1` in (iii), we get
`4x+2y-4z-6=0 rArr 2x+y-2z-3=0`.
Putting, `lambda=-1`, in (iii), we get
`-2x+4y+4z-6=0 rArr 2x+y-2z-3=0`.
Hence, the required equations of the planes are
`2x+y-2z-3 =0` and `x-2y-2z+3=0`.
Note: The vector equations of these planes are
`vecr.(2hati+hatj-2hatk)=3` and `vecr.(hati-2hatj-2hatk)+3=0`.