If the shortest distance between the lines ` \frac{x-1}{\alpha} = \frac{y+1}{-1} = \frac{z}{1} , (\alpha \ne -1)` and `

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Q 1684834757.     If the shortest distance between the lines ` \frac{x-1}{\alpha} = \frac{y+1}{-1} = \frac{z}{1} , (\alpha \ne -1)` and `x + y + z + 1 = 0 = 2x - y + z + 3` is `\frac{1}{\sqrt{3}}`, then `a` value of `\alpha` is :
JEE 2015 Mains 10 April
A

`\frac{32}{19}`

B

`- \frac{16}{19}`

C

`- \frac{19}{16`

D

`\frac{19}{32}`

HINT


Solution
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

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