Question

In $R^3$, consider the planes $P_1:y=0$ and $P_2:x+z=1$. Let $P_3$ be a plane, different from $P_1$ and $P_2$, which passes through the intersection of $P_1$ and $P_2$. If the distance of the point $(0,1,0)$ from $P_3$ is $1$ and the distance of a point $(\alpha ,\beta ,\gamma )$ from $P_3$ is $2$, then which of the following relations is (are) true?

• A. $2\alpha +\beta +2\gamma +2=0$
• B. $2\alpha -\beta +2\gamma +4=0$
• C. $2\alpha +\beta -2\gamma -10=0$
• D. $2\alpha -\beta +2\gamma -8=0$

Answer 1

Answer: (B), (D)

$2\alpha -\beta +2\gamma -8=0$
$2\alpha -\beta +2\gamma +4=0$

Using the concept of family of planes, the equation of $P_3$ can be written as $(x+z-1)+\lambda y=0$
The distance of $(0,1,0)$ from $P_3$ is $1$.
Hence,
$\frac{-1+\lambda }{\sqrt{{\lambda }^2+2}}=1$
On squaring ,
${\lambda }^2-2\lambda +1={\lambda }^2+2$
$\lambda =-\frac{1}{2}$
Hence, $P_3:2x-y+2z-2=0$
The distance of $(\alpha ,\beta ,\gamma )$ from $P_3$ is $2$.
Hence,
$\left|\frac{2\alpha -\beta +2\gamma -2}{3}\right|=2$
$\Rightarrow 2\alpha -\beta +2\gamma -8=0$
or
$2\alpha -\beta +2\gamma +4=0$
Hence, options B and D are correct.