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

Let \(P\) be the point of intersection of the line \(\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}\) and the plane \(x + y + z =2\) If the distance of the point \(P\) from the plane \(3 x-4 y+12 z=32\) is \(q\), then \(q\) and \(2 q\) are the roots of the equation

  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, \(x^2-18 x+72=0\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app