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

#### Solution

The equations of the given planes are

`vecr (hati+2hatj+3hatk)-4=0 ........(1)`

`vec r (2hati+hatj-hatk)+5=0 .........(2)`

The equation of the plane passing through the intersection of the planes (1) and (2) is

`[vecr (hati+2hatj+3hatk)-4]+lambda[ vec r (2hati+hatj-hatk)+5]=0`

`vecr[(1+2lambda)hati+(2+lambda)hatj+(3-lambda)hatk]=4-5lambda.............(3)`

Given, the plane (3) is perpendicular to the plane `vecr(5 hati+3hatj-6hatk)+8=0`

`(1+2lambda)xx5+(2+lambda)xx3+(3-lambda)xx(-6)=0`

`19lambda-7=0`

`lambda=7/19`

Putting `lambda=7/19 ` in (3), we get

`vec r[(1+14/19)hati+(2+7/19)hatj+(3-7/19)hatk]=4-35/19`

`vecr(33/19hati+45/19hatj+50/19hatk)=41/19`

`vecr(33hati+45hatj+50hatk)=41`

Thus, the equation of the required plane is

`vecr(33hati+45hatj+50hatk).=41`