Chapter 11 - Section 11.1 - Three-Dimensional Coordinate Systems - Exercises - Page 600: 62

$|PA|=|PB|=\sqrt 6$

The distance between two points can be calculated as: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ We need to verify that $|PA|=|PB|$ Thus, $|PA|=\sqrt{(3-2)^2+(1+1)^2+(2-3)^2}=\sqrt{6}$ and $|PB|=\sqrt{(3-4)^2+(1-3)^2+(2-1)^2}=\sqrt{6}$ Hence, the perimeter of the triangle is: $|PA|=|PB|=\sqrt 6$