Official solution
Plane through line of intersection is
x + y + z + 1 + \[\lambda \] (2x --y + z + 3) = 0
It should be parallel to given line \[{{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}\]
\[ \therefore \] 0(1 + 2\[\lambda \]) - 1(1 - \[\lambda \]) + 1(1 + \[\lambda \]) = 0 \[ \Rightarrow \] \[\lambda \] = 0
\[ \therefore \] Required Plane : x + y + z + 1 = 0
Shortest distance of (1, --1, 0) from this plane
= \[{{\left| {1 - 1 + 0 + 1} \right|} \over {\sqrt {{1^2} + {1^2} + {1^2}} }}\] = \[{1 \over {\sqrt 3 }}\]