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