Chapter 12 - Vectors and the Geometry of Space - 12.5 Equations of Lines and Planes - 12.5 Exercises - Page 873: 72

$\frac{40}{\sqrt {21}}$

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