Equation of line of projection of the line `3x-y+2z-1=0=x+2y-z=2` on the plane `3x+2y+z=0` is

Options

A. `(x+1)/11=(y-1)/-9=(z-1)/-15`
B. `3x-8y+7z+4=0=3x+2y+z`
C. `(x+12)/11 = (y+8)/-9 = (z+14)/15`
D. `(x+12)/11=(y+8)/-9=(z+14)/-15`

Answer

B

Explanation

Equation of a plane passing through the line
`3x-y+2z-1=0=x+2y-z-2` is `3x-y+2z-1+lambda(x+2y-z-2)=0`
`rArr (3+lambda)x+(-1+2lambda)y+(2-lambda)z+(-1-2lambda)`
=0
Since it must be perpendicular to the given plane.
`rArr (3+lambda)3+(-1+2lambda(x+2y-z-2)=0`
`rArr (3+lambda)x+(-1+2lambda)2+(2-lambda)1=0`
`rArr 6lambda-9 rArr lambda=-3//2`
`rArr` Plane is `3x-8y+7z+4=0`
Now line of projection is line of intersection of planes `3x-8y+7z+4=0` and `3x+2y+z=0`