JEE Math Practice Question
If the line, $${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + \lambda } \over { - 2}}$$ lies in the plane, 2x−4y+3z=2, then the shortest distance between this line and the line, $${{x - 1} \over {12}} = {y \over 9} = {z \over 4}$$ is :
- A.2
- B.1
- C.0
- D.3
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