The unit vector perpendicular to both $${L_1}$$ and $${L_2}$$ is :

Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

The shortest distance between $${L_1}$$ and $${L_2}$$ is :

Consider three points $$P = ( - \sin (\beta - \alpha ), - cos\beta ),Q = (cos(\beta - \alpha ),\sin \beta )$$ and $$R = (\cos (\beta - \alpha + \theta ),\sin (\beta - \theta ))$$ where $$0 < \alpha ,\beta ,\theta < {\pi \over 4}$$. Then :