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

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