Answer
$d=\frac{6}{\sqrt 14}$
Work Step by Step
Using the formula from Exercise 28, the distance of a point $P(x_0,y_0,z_0)$ to the plane $P..ax+by+cz+d=0$ is given by:
$$d(P,P)=\frac{|x_0.a+y_0.b+z_0.c+d|}{\sqrt a^2+b^2+c^2} \\
=\frac{0.2+0.(-1)+0.3+(-6)}{\sqrt 2^2+(-1)^2+3^2}\\
= \frac{|-6|}{\sqrt 4+1+9}\\
=\frac{6}{\sqrt 14}$$
Hence, $d=\frac{6}{\sqrt 14}$
