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

उस समतल का समीकरण, जो रेखाओं  \(\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}\) तथा \(\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}\) के प्रतिच्छेदन बिंदु से हो कर जाता है, तथा मूलबिंदु से अधिकतम दूरी पर है

  1. A \(7x +2y+4z= 54\)
  2. B \(3x+4y+ 5z= 49\)
  3. C \(4x+3y+5z = 50\)
  4. D \(5x+4y+3z = 57\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4x+3y+5z = 50\)

Step-by-step Solution

Detailed explanation

Given equation of lines are \(\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}\) .......\((1)\) and \(\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}\) ....\((2)\) Any point on line \(( 1)\) is \(P\) \((3 \lambda+1, \lambda+2,2 \lambda+3)\) and on line \(( 2)\) is \(Q\)…
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