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}`

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