## Thomas' Calculus 13th Edition

$3$
Apply the distance formula, $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$ $d=\sqrt{(3-1)^{2}+(3-1)^{2}+(0-1)^{2}}$ $d=\sqrt{4+4+1}=\sqrt{9}$ $d=3$