Answer
The distance from $Q = \left( {1,1,1} \right)$ to the plane $2x+y+5z=2$ is
$distance \simeq 1.0955$
Work Step by Step
By Theorem 1, the plane $2x+y+5z=2$ has normal vector ${\bf{n}} = \left( {2,1,5} \right)$.
The distance from $Q = \left( {{x_1},{y_1},{z_1}} \right)$ to the plane $\Pi $ is the distance to the point $P$ on $\Pi $ closest to $Q$. It is given by Eq. (8):
Distance from $Q$ to ${\Pi }$ $ = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} - d} \right|}}{{||{\bf{n}}||}}$
Using Eq. (8), the distance from $Q = \left( {1,1,1} \right)$ to the plane $2x+y+5z=2$ is
$distance = \frac{{\left| {2\cdot1 + 1\cdot1 + 5\cdot1 - 2} \right|}}{{||\left( {2,1,5} \right)||}}$
$distance = \frac{6}{{\sqrt {{2^2} + {1^2} + {5^2}} }} = \frac{6}{{\sqrt {30} }} \simeq 1.0955$
