Find the equation of the plane which contains the line of intersection of the planes `vecr.(hati+2hatj+3hatk)-4=0, vecr.(2hati+hatj-hatk)+5=0` and which is perpendicular to the plane `vecr.(5hati+3hatj-6hatk)+8=0`

Explanation

The equation of the given plane are
`vecr.(hati+2hatj+3hatk)=4` and `vecr.(-2hati-hatj+hatk)=5`.
These equations are of the form `vecr.(vecn_(1))=q_(1)` and `vecr.vecn_(2)=q_(2)`, where
`vecn_(1)=(hati+2hatj+3hatk), vecn_(2)=(-2hati-hatj+hatk), q_(1)=4` and `q_(2)=5`.
The equation of the plane through the intersections of the given planes is given by
`rArr vecr.[(hati+2hatj+3hatk)+lambda(-2hati-hatj+hatk)]=(3+5lamba)`
`rArr vecr.[(1-2lambda)hati+(2-lambda)hatj+(3+lambda)hatk]=(4+5lambda)`.........(i)
This plane (i) is perpendicular to the plane `vecr.(-5hati-3hatj+6hatk)=8`, so we have
`(1-2lambda) xx (-5) + (2-lambda) xx (-3) + (3+lambda) xx 6=0`
`rArr (-5+10lambda)+(-6+3lambda)+(18+6lambda)=0`
`rArr 19lambda=-7 rArr lambda=-7/19`.
Putting `lambda=-7/19` in (i), we get
`vecr.[(1+14/19)hati+(2+7/19)hatj+(3-7/19)hatk]=(4-35/19)`
`rArr vecr.(33hati+45hatj+50hatk)=41`.
Hence, the required equation of the plane is
`vecr.(33hati+45hatj+50hatk)=41`.