Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x − y + z = 0. Also find the distance of the plane, obtained above, from the origin.

#### Solution

Equation of the plane through the line of intersection of the two planes is

` (x+y+z−1)+λ(2x+3y+4z−5)=0⇒(1+2λ)x+(1+3λ)y+(1+4λ)z−1−5λ=0 .........(1)`

Normal vector of the required plane is

`vecN =(1+2λ)hati+(1+3λ)hatj+(1+4λ)hatk`

Since this plane is perpendicular to* x-y+z=0, *their normals are also perpendicular.

`⇒vecN_1.vecN_2=0`

`⇒((1+2λ)hati+(1+3λ)hatj+(1+4λ)hatk).(hati−hatj+hatk)=0`

`⇒1+2λ−1−3λ+1+4λ=0`

`⇒λ=−1/3`

Substituting the value of λ in equation (1), to obtain the required equation of the plane

`(1+2(−1/3))x+(1+3(−1/3))y+(1+4(−1/3))z−1−5(−1/3)`

=0

⇒x−z+2=0

To find out the distance of this plane from the origin we need to convert this equation to the standard form:

lx+my+nz=d

Dividing both sides of the equation by `sqrt2`

, we get:

`x/sqrt2−z/sqrt2=−2/sqrt2`

`x/sqrt2−z/sqrt2=−sqrt2`

Hence, the distance of the plane from the origin is `sqrt2` units.