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 (α, β,γ) drom $P_3$ is 2, then which of the following relations is (are) true ?

(a) $2α+ β + 2γ + 2= 0 $

(b) $2α- β + 2γ + 4= 0 $

(c) $2α+ β - 2γ -10= 0 $

(d) $2α- β + 2γ -8= 0 $

(b), (d)

The plane passing through the intersection of planes $P_1 : y = 0 $ and $P_2 : x + z = 1 $ is

$P_3 : (x + z - 1) + λy = 0 $ or, $x + λy + z - 1= 0 $$.......(i)

It is given that the distances of the points (0, 1, 0) and (α,β,γ) from (i) are 1 and 2 respectively.

$∴ \begin{vmatrix} \frac{0+λ+0-1}{\sqrt{1+λ^2 +1}}\end{vmatrix}=1 $ and $\begin{vmatrix} \frac{\alpha +λ\beta +\gamma -1}{\sqrt{1+λ^2 +1}}\end{vmatrix}=2 $

$⇒ \frac{|λ-1|}{\sqrt{λ^2+2}}=1 $ and $\frac{|\alpha + λ\beta + \gamma - 1|}{\sqrt{λ^2+2}}=2$

$⇒ (λ-1)^2 = λ^2 + 2 $ and $|\alpha + λ\beta + \gamma -1| = 2 \sqrt{λ^2 + 2}$

$⇒ λ= -\frac{1}{2}$ and $|\alpha + λ\beta + \gamma -1|= 2 \sqrt{λ^2 + 2}$

$⇒ |2\alpha - \beta + 2 \gamma - 2| = 6 $

$⇒ 2\alpha - \beta + 2 \gamma - 2= ±6⇒2\alpha - \beta + 2\gamma - 8 = 0 $

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