#### Answer

$d = 3$

#### Work Step by Step

$P_1 = (1,-1,1)$
$P_2 = (2,1,3)$
We can find the distance between these two points:
$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
$d = \sqrt{(2-1)^2+(1-(-1))^2+(3-1)^2}$
$d = \sqrt{(1)^2+(2)^2+(2)^2}$
$d = \sqrt{1+4+4}$
$d = \sqrt{9}$
$d = 3$