Question

The equation of the plane whose distance from the origin is unity and which passes through the line $x+3y+6=0=3x-y-4z$ is

• A. $x+2y+2z-3=0$
• B. $2x+y+2z-3=0$
• C. $2x+y-2z+3=0$
• D. $x-2y-2z-3=0$

Answer 1

Answer: (C), (D)

The correct options are
C $2x+y-2z+3=0$
D $x-2y-2z-3=0$

The equation of any plane which contains the given line is $x+3y+6+\lambda (3x-y-4z)=0$
i. e., $(1+3\lambda )x+y(3-\lambda )-4\lambda z+6=0$
If the distance of this plane from the origin is unity, then
$\frac{6}{\sqrt{(1+3\lambda {)}^2+(3-\lambda {)}^2+16{\lambda }^2}}=1$
$\Rightarrow (1+3\lambda {)}^2+(3-\lambda {)}^2+16{\lambda }^2=36$
$\Rightarrow 26{\lambda }^2=26\Rightarrow \lambda =\pm 1$
The equation of the planes are $2x+y-2z+3=0$ (for $\lambda =1$)
and $x-2y-2z-3=0$ (for $\lambda =-$1)
Hence, options C and D are correct.