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

If the distance between the plane, \(23 \mathrm{x}-10 \mathrm{y}-2 \mathrm{z}+48=0\) and the plane containing the lines \(\frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3}\) and \(\frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda}(\lambda \in R)\) is equal to \(\frac{\mathrm{k}}{\sqrt{633}},\) then \(\mathrm{k}\) is equal to

  1. A \(2\)
  2. B \(3\)
  3. C \(6\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

If \(\lambda=-7,\) then planes will be parallel \& distance between them will be \(\frac{3}{\sqrt{633}} \Rightarrow \mathrm{k}=3\) But if \(\lambda \neq-7,\) then planes will be intersecting and distance between them will be \(0\)
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