Equations of the line passing through `(1,1,1)` and perpendicular to the plane `2x+3y+z+5=0` are

Options

A. `(x-1)/1=(y-1)/2=(z-1)/1`
B. `(x-1)/(-1)=(y-1)/1=(z-1)/(-1)`
C. `(x-1)/3=(y-1)/2=(z-1)/1`
D. `(x-1)/2=(y-1)/3=(z-1)/1`

Answer

B

Explanation

Clearly, all lines in the given optics pass through (1,1,1). We know that a line is parallel to a plane if the normal to the plane is perpendicular to the line.
We obsere that the vector normal to the given plane is `vecn=2hati+3hatj+hatk` and the line in option b is parallel to the vector `vecb=-hati+hatj-hatk` such that
`vecb.vecn=-2+3-1=0` i.e. `vecb_|_vecn`.
Hence option b is correct.