Chapter 9 - Polar Coordinates; Vectors - 9.6 Vectors in Space - 9.6 Assess Your Understanding - Page 625: 18

$2\sqrt{19}$

The distance $d$ from $P_1(x_1,y_1,z_1)$ to $P_2(x_2,y_2,z_2)$ is given by the formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. Hence here $d=\sqrt{(4-(-2))^2+(0-2)^2+(-3-3)^2}=\sqrt{6^2+(-2)^2+(-6)^2}=\sqrt{36+4+36}=\sqrt{76}=2\sqrt{19}$