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 planes are
`" "vecr*(hati+2hatj+3hatk)-4=0" "`(i)
`" "vecr*(2hati+hatj-hatk)+5=0" "` (ii)
The equation of the plane passing through the line of intersection of the plane given in (i) and (ii) is
`" "[vecr*(veci+2vecj+3veck)-4]+lamda[vecr*(2hati+hatj-hatk)+5]=0`
`" "vecr*[(2lamda+1)hati+(lamda+2)hatj+(3-lamda)hatk]+(5lamda-4)=0" "`(iii)
The plane in (iii) is perpendicular to the plane, `vecr*(5hati+3hatj-6hatk)+8=0`. Thus,
`" "5(2lamda+1)+3(lamda+2)-6(3-lamda)=0`
or `" "19 lamda-7=0`
or `" "lamda=(7)/(19)`
Substituting `lamda=(7)/(19)` in (iii), we get
`" "vecr*[(33)/(19)hati+(45)/(19)hatj+(50)/(19)hatk]-(41)/(19)=0`
or `" "vecr*(33hati+45hatj+50hatk)-41=0`
or `" "33x+45y+50z-41=0`