The equation of a plane through the point A(1,0,-1) and perpendicular to the line `(x+1)/2=(y+3)/4=(z+7)/-3` is

Options

A. `2x+4y-3z=3`
B. `2x-4y+3z=5`
C. `2x+4y-3z=5`
D. `x+3y+7z=-6`

Answer

c

Explanation

D.r.'s of the given line are 2,4,-3.
Let the required equation of plane through (1,0,-1) be
`a(x-1)+b(y-0)+c(z-1)=0`.
Since, the line is perpendicular to the plane, so it is parallel to the normal to the plane.
`2(x-1)+4(y-0)-3(z+1)=0 rArr 2x+4y-3z=5.`