Equation of line passing through A(1,0,3), intersecting the line `(x/2=(y-1)/3=(z-2)/1)` and parallel to the plane `x+y+z=2` is

Options

A. `(3x-1)/2=(2y-3)/3=(2z-5)/-1`
B. `(x-1)/2=(y-0)/3=(z-3)/-1`
C. `(x-(2//3))/1=(y-(3//2))/0=(z+(1//2))/3`
D. `(3x-1)/2=(2y-3)/-3=(6z-13)/5`

Answer

D

Explanation

Any points on the line `x/2=(y-1)/3=(z-2)/1` is
`B(2lambda,1+3lambda,2+lambda)`
Now, AB is parallel to the plane `x+y+z=2`, where `A-=(1,0,3)`.
`therefore (2lambda-1)+(1+3lambda)+(lambda-1)=0`
`rArr lambda=1/6`
`therefore` Point B `-=(1/3,3/2,13/6)`
Direction ratio's of AB are
`1/3-1,3/2-0,13/6-3-=-2/3,3/2,-5/6`
`therefore` Its equation is
`(x-(1//3))/(-2//3) = (y-(3//2))/(3//2) = (z-(13//6))/(-5//6)`
or `(3x-1)/2=(2y-3)/(-3) = (6z-13)/5`