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

Explanation

Given planes are
`vecr.(hati+2hatj+3hatk) = 0 "….."(1)`
`vecr.(2hati+hatj-hatk) + 5 = 0 "……"(2)`
and `vecr.(5hati+3hatj-6hatk) + 8 =0"….."(3)`
Equation of plane through the intersection of plane (1) and (2) is
`[vecr.(hati+2hatj+3hatk)-4]+lambda[vecr.(2hati+hatj-hatk)+5]=0`
`rArr [vecr.(1+2lambda)hati+(2+lambda)hatj+(3-lambda)hatk)] +(5lambda-4)=0 "....."(1)`
This plane is perpendicular to plane (3).
`:. 5(1+2lambda)+3(2+lambda)-6(3-lambda)=0`
`rArr 19lambda = 7 rArr lambda = (7)/(19)`
Put the value of `lambda` in equation (4),
`vecr.[(1+(14)/(19))hati+(2+(7)/(19))hatj+(3-(7)/(19))hatk]`
`+(35)/(19)-4=0`
`rArr vecr.[(33/19)hati+(45/19)hatj+(50/19)hatk]-41/19=0`
`rArr r.(33hati+45hatj+50hatk)-41 = 0 "....."(5)`
which is the vector equation of required plane.
Put `r = x hati+yhatj+zhatk` in equation (5), the cartesian equation of plane is
`(xhati+yhatj+zhatk).(33hati+45hatj+50hatk)-41 = 0`
`rArr 33x+45y+50z-41 = 0`