## University Calculus: Early Transcendentals (3rd Edition)

$2 \sqrt 3$
Let $A=(0,0,0)$ and $B=(2,-2,-2)$ 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 $|AB|=\sqrt{(2-0)^2+(-2-0)^2+(-2-0)^2}=\sqrt {4+4+4}=\sqrt{12}=2 \sqrt 3$