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

If the shortest distance between the lines \(\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}\) and \(\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{z+2 \sqrt{6}}{5}\) is 6 , then sum of squares of all possible values(s) of \(\lambda\) is

  1. A 381
  2. B 384
  3. C 322
  4. D 386
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(B) 384

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