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
- A 381
- B 384
- C 322
- D 386
Answer & Solution
Correct Answer
(B) 384
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