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JEE Main 2018Medium

Q.The shortest distance between the line x+y+z=3x + y + z = 3 and 2x+3y+4z+5=02x + 3y + 4z + 5 = 0 is

a
1/291/\sqrt{29}
b
3/293/\sqrt{29}
c
5/295/\sqrt{29}
d
7/297/\sqrt{29}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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