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

माना रेखाएँ \(l_1: \frac{\mathrm{x}+5}{3}=\frac{\mathrm{y}+4}{1}=\frac{\mathrm{z}-\alpha}{-2}\) तथा \(l_2: 3 \mathrm{x}+2 \mathrm{y}+\mathrm{z}-2=0=\mathrm{x}-3 \mathrm{y}+2 \mathrm{z}-13\) सहलतीय हैं। यदि \(l_1\) पर बिंदु \(\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})\), बिंदु \(\mathrm{Q}(-4,-3,2)\) के निकटतम है, तो \(|\mathrm{a}|+|\mathrm{b}|+|\mathrm{c}|\) बराबर है

  1. A \(12\)
  2. B \(14\)
  3. C \(10\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(10\)

Step-by-step Solution

Detailed explanation

\((3 x+2 y+z-2)+\mu(x-3 y+2 z-13)=0\) \(3(3+\mu)+1 \cdot(2-3 \mu)-2(1+2 \mu)=0\) \(9-4 \mu=0\) \(\mu=\frac{9}{4}\) \(4(-15-8+\alpha-2)+9(-5+12+2 \alpha-13)=0\) \(-100+4 \alpha-54+18 \alpha=0\) \(\Rightarrow \alpha=7\)…
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