Let R3 , consider the plane P1 : y = 0 and P2 : x + z = 1. Let P3 be a 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?

Equation of any plane through intersection of P1 andP2 isP3 : (x + z –1) + ky = 0Distance of P3 from (0, 1, 0) is|−1+k|1+k2+1=1⇒(K−1)2=K2+2⇒K2−2K+1=K2+2⇒K=−12∴ Equation of P3 is (x+z−1)−12y=0 or 2x−y+2z−2=0As distance of (1) from (α,β,γ) is 2|2α−β+2γ−2|4+1+4=2⇒2α−β+2γ−2±6⇒2α−β+2γ+4=0 or 2α−β+2γ−8=0

Let R3 , consider the plane P1 : y = 0 and P2 : x + z = 1. Let P3 be a 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?

Equation of any plane through intersection of P1 andP2 isP3 : (x + z –1) + ky = 0Distance of P3 from (0, 1, 0) is|−1+k|1+k2+1=1⇒(K−1)2=K2+2⇒K2−2K+1=K2+2⇒K=−12∴ Equation of P3 is (x+z−1)−12y=0 or 2x−y+2z−2=0As distance of (1) from (α,β,γ) is 2|2α−β+2γ−2|4+1+4=2⇒2α−β+2γ−2±6⇒2α−β+2γ+4=0 or 2α−β+2γ−8=0

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Let R³ , consider the plane P₁ : y = 0 and P₂ : x + z = 1. Let P₃ be a 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?

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$2 α + β - 2 γ - 10 = 0$

$2 α + β + 2 γ + 2 = 0$

$2 α - β + 2 γ - 2 = 0$

$2 α - β + 2 γ + 4 = 0$

answer is B, D.

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Detailed Solution

Equation of any plane through intersection of P₁ and
P₂ is
P₃ : (x + z –1) + ky = 0
Distance of P3 from (0, 1, 0) is
$\begin{array}{l} \frac{| - 1 + k |}{\sqrt{1 + k^{2} + 1}} = 1 \Rightarrow (K - 1)^{2} = K^{2} + 2 \\ \Rightarrow K^{2} - 2 K + 1 = K^{2} + 2 \Rightarrow K = - \frac{1}{2} \\ ∴ Equation of P_{3} is (x + z - 1) - \frac{1}{2} y = 0 \\ or 2 x - y + 2 z - 2 = 0 \end{array}$
As distance of (1) from $(α, β, γ)$ is 2
$\begin{array}{l} \frac{| 2 α - β + 2 γ - 2 |}{\sqrt{4 + 1 + 4}} = 2 \\ \Rightarrow 2 α - β + 2 γ - 2 \pm 6 \Rightarrow 2 α - β + 2 γ + 4 = 0 \\ or 2 α - β + 2 γ - 8 = 0 \end{array}$