The plane \(\Pi _ { 1 }\) contains the point \(( 3,3 , - 2 )\) and the line \(\frac { x - 1 } { 2 } = \frac { y - 2 } { - 1 } = \frac { z + 1 } { 4 }\)
Show that a cartesian equation of the plane \(\Pi _ { 1 }\) is
$$3 x - 10 y - 4 z = - 13$$
The plane \(\Pi _ { 2 }\) is parallel to the plane \(\Pi _ { 1 }\)
The point ( \(\alpha , 1,1\) ), where \(\alpha\) is a constant, lies in \(\Pi _ { 2 }\)
Given that the shortest distance between the planes \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\) is \(\frac { 1 } { \sqrt { 5 } }\)