Answer
$\frac{40}{\sqrt {21}}$
Work Step by Step
Given: $(-6,3,5)$ ; $x-2y-4z=8$
Distance formula between the point and the plane is given as:
$D=\frac{|ax+by+cz+d|}{\sqrt {a^2+b^2+c^2}}$
$D=\frac{|1 \cdot (-6)-2 \cdot 3+(-4) \cdot (5)-8|}{\sqrt {1^2+(-2)^2+(-4)^2}}$
$=\frac{40}{\sqrt {21}}$
