Question

The equation of the plane containing the lines $2x-y+z-3=0,3x+y+z=5$ and at a distance of $\frac{1}{\sqrt{6}}$ from the point $(2,1,-1)$ is

• A. $x+y+z-3=0$
• B. $2x-y-z-3=0$
• C. $2x-y+z-3=0$
• D. $62x+29y+19z-105=0$

Answer 1

Answer: (C), (D)

The correct options are
C $2x-y+z-3=0$
D $62x+29y+19z-105=0$

Equation of the plane is $2x-y+z-3+\lambda (3x+y+z-5)=0$ ... (i)

$\Rightarrow (2+3\lambda )x+(\lambda -1)y+(\lambda +1)z-(5\lambda +3)=0$

Its distance from $(2,1,-1)$ is $\frac{1}{\sqrt{6}}$.

$\Rightarrow \frac{|4+6\lambda +\lambda -1-\lambda -1-5\lambda -3|}{\sqrt{(2+3\lambda {)}^2+(\lambda -1{)}^2+(\lambda +1{)}^2}}=\frac{1}{\sqrt{6}}\Rightarrow 6(\lambda -1{)}^2=11{\lambda }^2+12\lambda +6$

$\Rightarrow (5\lambda +24)\lambda =0$
$\Rightarrow \lambda =0,-\frac{24}{5}$

If $\lambda =0$, then equation of the plane is $2x-y+z-3=0$

If $\lambda =\frac{-24}{5}$, then equation of the plane is $5(2x-y+z-3)-24(3x+y+z-5)=0$

$\Rightarrow 62x+29y+19z-105=0$