__Ask a Question__

### Question Asked by a Student from EXXAMM.com Team

** 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**