The equation of passing through line `(x-1)/(2)=(y+1)/(-1)=(z-3)/(4)` and perpendicular to the plane `x+2y+z=12` is given by `ax+by+cz+4=0`, then

Options

A. `a=-8,b=2,c=-5`
B. `a=-9,b=-2,c=-5`
C. `a=9,b=-2,c=-5`
D. None of these

Answer

C

Explanation

Equation of any plane passing through given line is
`a(x-1)+b(y+1)+c(z-3)=0` . . . (i)
Above plane is perpendicular to the plane
`x+2y+z=12`
`:." "a+2b+c=0" "[becausea_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0]`
Also, normal to the plane is perpendicular to the line
`:." "2a-b+4c=0`
`rArr" "(a)/(8+1)=(b)/(2-4)=(c)/(-1-4)`
`rArr" "(a)/(9)=(b)/(-2)=(c)/(-5)`
`:.9(x-1)-2(y+1)=5(z-3)=0`
`rArr" "9x-2y-5z+4=0`
`:." "a=9,b=-2,c=-5`