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

मान लीजिए कि रेखाओं \(\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}\) और \(\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}\) के प्रतिच्छेदन बिंदु की बिंदु \((7,8,9)\) से दूरी \(d\) है। तब \(\mathrm{d}^2+6\) = ...........

  1. A \(72\)
  2. B \(69\)
  3. C \(75\)
  4. D \(78\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(75\)

Step-by-step Solution

Detailed explanation

\( \frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda \) .................(\(1\)) \( x=3 \lambda-6, y=2 \lambda, z=\lambda-1 \) \( \frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu \) ........................(\(2\)) \( x=4 \mu+7, y=3 \mu+9, z=2 \mu+4 \)…
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