The equation of a plane passing through the line of intersection of the planes x+2y+3z = 2 and x - y + z = 3 and at a distance `2/sqrt 3` from the point (3, 1, -1) is

Options

A. `5x-11y+z=17`
B. `sqrt(2)x+y=3sqrt(2)-1`
C. `x+y+z=sqrt(3)`
D. `x-sqrt(2)y=1-sqrt(2)`

Answer

A

Explanation

The equation of a plane passing through the line of intersection of the planes `x+2y+3z=2` and `x-y+z=3` is
`(x+2y+3z-2)+lamda(x-y+3)=0`
`impliesx(lamda+1)+y(2-lamda)+z(3+lamda)-2-3lamda=0` …………i
It is at a distance of `2//sqrt(3)` units from the point `(3,1,-1)`
`:.(|3(lamda+1)+(2-lamda)-(3+lamda)-2-3lamda|)/(sqrt((lamda+1)^(2)+(2-lamda)^(2)+(3+lamda)^(2)))=2/(sqrt(3))`
`implies (|-2lamda|)/(sqrt(3lamda^(2)+4lamda+14))=2/(sqrt(3))`
`implies3lamda^(2)+4lamda+14=3lamda^(2)implies4lamda=-14implieslamda=(-7)/2`
Putting `lamda=-7/2` in i we obtain that the equation of the required plane is
`-5/2 x +11/2 y -z/2+17/2=0` or `5x-11y+z-17=0`