Chapter 12 - Section 12.1 - Three-Dimensional Coordinate Systems - 12.1 Exercises - Page 835: 10

$d=\sqrt{138}$

Given two points, $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ , the distance between these two points is given by $d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. In our case, $P_1(-6,-3,0)$ and $P_2(2,4,5)$. Substituting in our formula, we end up with: $$d=\sqrt{(2-(-6))^2 + (4-(-3))^2 + (5-0)^2} = \sqrt{138}$$