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)

The correct options are
B $2\alpha -\beta +2\gamma +4=0$
D $2\alpha -\beta +2\gamma -8=0$

Equation of family of planes passing through $P_1$ and $P_2$ (plane $P_3$) is given by: $P_2+\lambda P_1=0$
$\Rightarrow x+\lambda y+z-1=0$

The distance of the point $(0,1,0)$ from $P_3$ is $1.$
$\Rightarrow \frac{|\lambda -1|}{\sqrt{{\lambda }^2+2}}=1$
$\Rightarrow \lambda =-\frac{1}{2}$

Thus, the equation of the plane $P_3$ is $2x-y+2z-2=0$
Now, the distance of a point $(\alpha ,\beta ,\gamma )$ from $P_3$ is $2.$
$\Rightarrow \frac{|2\alpha -\beta +2\gamma -2|}{3}=2$
$\Rightarrow 2\alpha -\beta +2\gamma -8=0\text{or}2\alpha -\beta +2\gamma +4=0$