## Thomas' Calculus 13th Edition

$\sqrt 6$
Formula to find the distance between two points is $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ We will have to verify that $|PA|=|PB|$ $|PA|=\sqrt{(3-2)^2+(1+1)^2+(2-3)^2}=\sqrt{(1)^2+(2)^2+(-1)^2}=\sqrt{6}$ and $|PB|=\sqrt{(3-4)^2+(1-3)^2+(2-1)^2}=\sqrt{(-1)^2+(-2)^2+(1)^2}=\sqrt{6}$ Hence, $|PA|=|PB|=\sqrt 6$