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

જો રેખાઓ \(\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}\) અને \(\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}\) વચ્ચેનું ન્યુનતમ અંતર \(\frac{6}{\sqrt{5}}\) હોય, તો \(\lambda\) ની શક્ય તમામ કિમતોનો સરવાળો ........... થાય.

  1. A \(5\)
  2. B \(8\)
  3. C \(7\)
  4. D \(10\)
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Answer & Solution

Correct Answer

(B) \(8\)

Step-by-step Solution

Detailed explanation

\( \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \) \( \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}\) the shortest distance between the lines…
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