Question

Let plane $P=0$ pass through the intersection of planes $2x-y+z-3=0$ and $3x+y+z-5=0$. If the distance of plane $P=0$ from $(2,1,-1)$ is $\frac{1}{\sqrt{6}}$ then its actuation can be:

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

Answer 1

The correct option is B $62x+29y+19z-105=0$

Equation of the plane will be of the form $(2x-y+z-3)+\lambda (3x+y+z-5)=0$

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

The distance from $(2,1,-1)$ to this plane is given as $\frac{1}{\sqrt{6}}$

$\therefore \frac{2(3\lambda +2)+\lambda -1-\lambda -1-5\lambda -3}{\sqrt{(3\lambda +2{)}^2+(\lambda -1{)}^2+(\lambda +1{)}^2}}=\frac{1}{\sqrt{6}}$

$\frac{(\lambda -1{)}^2}{9{\lambda }^2+12\lambda +4+2{\lambda }^2+2}=\frac{1}{6}$

$6({\lambda }^2-2\lambda +1)=9{\lambda }^2+2{\lambda }^2+12\lambda +6$

$6{\lambda }^2-12\lambda +6=11{\lambda }^2+12\lambda +6$

$5{\lambda }^2+24\lambda +0=0$

$\lambda =0,-\frac{24}{5}$

$x\left(\frac{-62}{5}\right)+y\left(\frac{-29}{5}\right)+z\left(\frac{-19}{5}\right)=-21$

or

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