The equation of the plane through the intersection of the planes `vecr.(2hati+6hatj)+12=0` and `vecr.(3hati-hatj+4hatk)=0` and at a unit distance from the origin, is

Explanation

The equations of the given planes are
`vecr.(2hati+6hatj)+12=0`
`rArr (xhati+yhatj+zhatk).(2hati+6hatj)+12=0`
`rArr 2x+6y+12=0 rArr x+3y+6=0` …………(i)
and `vecr.(3hati-hatj+4hatk)=0`
`rArr (xhati+yhatj+zhatk).(3hati-hatj+4hatk)=0`
`rArr 3x-y+4z=0`. .............(ii)
Equation of any plane through the intersection of these planes is
`(x+3y+4)+lambda(3x-y+4z)=0` for some real number `lambda`.
`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` or `lambda=-1`.
Putting `lambda=1`in (iii), we get
`4x+2y+4z+6=0 rArr 2x+y+2z-3=0`.
Putting `lambda=-1`. (iii), we get
`-2x+4y-4z+6=0 rArr x-2y+2z-3=0`.
Putting `lambda=-1` in (iii), we get
`-2x+4y-4z+6=0 rArr x-2y+2z-3=0`.
Hence, the required equations of the planes in Cartesian form are
`2x+y+2z+3=0` and `x-2y+2z-3=0`.
The vector forms of these equations are
`vecr.(hati+hatj+2hatk)+3=0` and `vecr.(hati-2hatj+2hatk)=3`.