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? $P_1:\ y = 0, \qquad P_2:\ x + z = 1$ $P_3 \text{ passes through } P_1 \cap P_2, \quad P_3 \neq P_1, P_2$

$\text{dist}\big((0,1,0), P_3\big) = 1, \qquad \text{dist}\big((\alpha,\beta,\gamma), P_3\big) = 2$

$\text{Find } P_3, \text{ then test which of (1)–(4) hold}$ $P_1:\ y = 0 \ (\text{xz-plane}), \qquad P_2:\ x + z = 1$ $P_1 \cap P_2:\ y = 0,\ x + z = 1 \quad (\text{a line})$ $P_3:\ (x + z - 1) + k\,y = 0$ $\Rightarrow\ x + ky + z - 1 = 0, \qquad \vec n = (1,\ k,\ 1)$ $\dfrac{|0 + k(1) + 0 - 1|}{\sqrt{1 + k^2 + 1}} = 1$ $\dfrac{|k - 1|}{\sqrt{k^2 + 2}} = 1$ $(k - 1)^2 = k^2 + 2$ $k^2 - 2k + 1 = k^2 + 2 \Rightarrow -2k = 1 \Rightarrow k = -\dfrac{1}{2}$ $P_3:\ x - \tfrac{1}{2}y + z - 1 = 0$ $2x - y + 2z - 2 = 0, \qquad |\vec n| = \sqrt{4 + 1 + 4} = 3$ $\dfrac{|2\alpha - \beta + 2\gamma - 2|}{3} = 2$ $|2\alpha - \beta + 2\gamma - 2| = 6$ $2\alpha - \beta + 2\gamma - 2 = 6 \Rightarrow 2\alpha - \beta + 2\gamma - 8 = 0 \ \text{(4)}$ $2\alpha - \beta + 2\gamma - 2 = -6 \Rightarrow 2\alpha - \beta + 2\gamma + 4 = 0 \ \text{(2)}$ $\boxed{\text{True relations: } (2) \text{ and } (4)}$ $(1), (3): \ \beta \text{ sign and } \gamma \text{ coefficient do not match}$ $P_3:\ 2x - y + 2z - 2 = 0, \quad \text{true: } (2), (4)$