Show that the points $P(-2, 3, 5)$,$Q(1, 2, 3)$,and $R(7, 0, -1)$ are collinear.

Points are collinear if they lie on the same straight line.
First,we calculate the distances between the points using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$PQ = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$.
$QR = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}$.
$PR = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2} = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$.
Since $PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = PR$,the sum of the lengths of two segments equals the length of the third segment.
Therefore,the points $P$,$Q$,and $R$ are collinear.