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

माना समतल \(P\) जिस पर रेखा \(x+10=\frac{8-y}{2}=z\) स्थित है, का समीकरण \(\mathrm{ax}+\mathrm{by}+3 \mathrm{z}=2(\mathrm{a}+\mathrm{b})\) है एवं बिन्दु \((1,27,7)\) से समतल \(\mathrm{P}\) की दूरी \(\mathrm{c}\) है। तब \(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\) is बराबर है____________. 

  1. A \(354\)
  2. B \(353\)
  3. C \(355\)
  4. D \(35.5\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(355\)

Step-by-step Solution

Detailed explanation

The line \(\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z}{1}\) have a point \((-10,8,0)\) with d. r. \((1,-2,1)\) \(\because\) the plane \(a x+b y+3 z=2(a+b)\) \(\Rightarrow b =2 a\) and dot product of d.r.'s is zero \(\therefore a-2 b+3=0\) \(\therefore a =1\) and \(b =2\) Distance…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app