Chapter 11 - Section 11.1 - Three-Dimensional Coordinate Systems - Exercises - Page 600: 41

$3$

Let $A=(1,1,1)$ and $B=(3,3,0)$ be the two given points. The distance between two points can be calculated as: $|AB|=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Thus, we have $|AB|=\sqrt{(3-1)^2+(3-1)^2+(0-1)^2}=\sqrt 9= 3$