In $R^{3}$, consider the planes $P_1 : y = 0$ and $P_2 : x + z = 1$. Let $P_3$ be the 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?

In R³, consider the planes P₁ : y = 0 and P₂ : x + z = 1. Let P₃ be the plane, different from P₁ and P₂, which passes through the intersection of P₁ and P₂. If the distance of the point (0, 1, 0) from P₃ is 1 and the distance of a point (α, β, γ) from P₃ is 2, then which of the following relations is (are) true?

$P_1 : y = 0$

$P_2 : x + z = 1$

$P_3 \supset P_1 \cap P_2, \qquad P_3 \neq P_1,\ P_3 \neq P_2$

$\text{To find: which relation(s) hold for } (\alpha, \beta, \gamma),$

$\text{given } d\big((0,1,0), P_3\big) = 1 \text{ and } d\big((\alpha,\beta,\gamma), P_3\big) = 2.$

$\text{Planes } P_1, P_2 \text{ intersect in a line } L$

$L = P_1 \cap P_2 : \quad y = 0, \quad x + z = 1$

$L:\ (x,y,z) = (1,0,0) + t(1,0,-1)$

$Q = (0,1,0), \quad \text{given } d(Q, P_3) = 1$

$1.\ \text{write the family of planes through } L$

$2.\ \text{use } d((0,1,0), P_3) = 1 \text{ to fix } P_3$

$3.\ \text{use } d((\alpha,\beta,\gamma), P_3) = 2 \text{ to test options}$

$P_3:\ (x + z - 1) + k\,y = 0$

$k = 0 \Rightarrow P_2 \quad \Rightarrow \quad k \neq 0$

$P_3 \neq P_1 \text{ is automatic here}$

$P_3:\ x + k\,y + z - 1 = 0$

$\vec{n} = (1,\ k,\ 1), \qquad |\vec{n}| = \sqrt{k^2 + 2}$

$d(Q, P_3) = \dfrac{|0 + k\cdot 1 + 0 - 1|}{\sqrt{k^2+2}} = \dfrac{|k-1|}{\sqrt{k^2+2}} = 1$

$(k-1)^2 = k^2 + 2 \ \Rightarrow \ k^2 - 2k + 1 = k^2 + 2$

$-2k = 1 \ \Rightarrow \ k = -\dfrac{1}{2}$

$P_3:\ x + z - 1 - \dfrac{1}{2}\,y = 0$

$\Rightarrow\ 2x - y + 2z - 2 = 0, \qquad |\vec{n}| = 3$

$d\big((\alpha,\beta,\gamma), P_3\big) = \dfrac{|2\alpha - \beta + 2\gamma - 2|}{3} = 2$

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

$+\ :\ 2\alpha - \beta + 2\gamma - 8 = 0 \quad \text{(option 4)}$

$-\ :\ 2\alpha - \beta + 2\gamma + 4 = 0 \quad \text{(option 2)}$

$\text{Options (2) and (4) are true}$

$\boxed{\text{Answer: (2), (4)}}$