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JEE Math Practice Question

If the shortest distance between the lines $$\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$$ and $$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$$ is $$\frac{13}{\sqrt{29}}$$, then a value of $$\lambda$$ is :

  1. A.$$\frac{13}{25}$$
  2. B.1
  3. C.$$-$$1
  4. D.$$-\frac{13}{25}$$

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