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JEE Mains · Maths · STD 12 - 11. three dimension geometry

माना रेखा \(\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}\) तथा समतल \(\mathrm{x}+\mathrm{y}+\mathrm{z}=2\) का प्रतिच्छेदन बिंदु \(\mathrm{P}\) है। यदि बिंदु \(3 x-4 y+12 z=32\) से दूरी \(q\) है, तो \(q\) तथा \(2 q\) किस समीकरण के मूल हैं ?

  1. A \(x ^2-18 x -72=0\)
  2. B \(x ^2+18 x +72=0\)
  3. C \(x ^2-18 x +72=0\)
  4. D \(x^2+18 x-72=0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(x ^2-18 x +72=0\)

Step-by-step Solution

Detailed explanation

\(P =(3 \lambda-3, \lambda-2,1-2 \lambda)\) \(P\) lies on the plane, \(x+y+z=2\) \(\Rightarrow \lambda=3\) \(P =(6,1,-5)\) \(q=\left|\frac{18-4-60-32}{\sqrt{9+16+144}}\right|=\frac{78}{13}=6\) \(q=6,2 q=12\) Equation, \(x^2-18 x+72=0\)
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