Determine the direction cosines of the normal to the plane and the distance
from the origin 2x + 3y - z = 5.

Answer

The direction ratios of the normal to the given plane are 2, 3, -1
The direction cosines of the normal are `(2)/sqrt(4+9+1), (3)/sqrt(4+9+1), (-1)/sqrt(4+9+1), i.e., (2)/sqrt14, (3)/sqrt14, (-1)/sqrt14 `
The distance of the plane from the origin = `abs((5)/sqrt(4+9+1)) = (5)/sqrt14`