Equation of the plane containing the straight line `x/2=y/3=z/4` and perpendicular to the plane containing the straight lines `x/2=y/4=z/2` and `x/4=y/2=z/3` is

Options

A. x+2y-2z=0
B. 3x+2y-2z=0
C. x-2y+z=0
D. 5x+2y-4z=0

Answer

C

Explanation

Equation of plane containing the line `(x)/(2)=(y)/(3)=(z)/(4)` is
`a(x-0)+b(y-0)+c(z-0)=0` . . . (i)
and `2a+3b+4c=0` . . (ii)
Another equation of plane containing the other two lines is
`a_(1)(x-0)+b_(1)(y-0)+c_(1)(z-0)=0` . . . (iii)
Also, `3a_(1)+4b_(1)+2c_(1)=0`
and `4a_(1)+2b_(1)+3c_(1)=0`
On solving, we get
`(a_(1))/(8)=(b_(1))/(-1)=(c)/(-10)`
`therefore` Eq. (iii) becomes
`8x-y-10c=0` . . . (iv)
Since, the plane (i) is perpendicular to the plane (ii)
`therefore8a-b-10c=0` . . . (v)
On solving Eqs. (ii) and (v), we get
`(a)/(-26)=(b)/(52)=(c_(1))/(-26)`
or `(a)/(1)=(b)/(-2)=(c)/(1)`
`therefore` From eq (i)
`x-2y+z=0`