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

રેખાઓ \(\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}\) અને \(\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}\) બિંદુ  \(\mathrm{P}\) આગળ છેદે છે. જે \(\mathrm{P}\) નું રેખા \(\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}\) થી અંતર \(l\) હોય, તો \(14 l^2 =\) ............

  1. A \(108\)
  2. B \(107\)
  3. C \(109\)
  4. D \(100\)
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Answer & Solution

Correct Answer

(A) \(108\)

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Detailed explanation

\( \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-7}{8}=\lambda \) \( \frac{\mathrm{x}+3}{4}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k} \) \( \Rightarrow \lambda+2=4 \mathrm{k}-3 \) \( -\lambda=3 \mathrm{k}-2 \)…
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