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

Let the points $(-2, 3, 5), (1, 2, 3),$ and $(7, 0, -1)$ be denoted by $P, Q,$ and $R$ respectively.
Points $P, Q,$ and $R$ are collinear if they lie on the same straight line.
$PQ = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$
$QR = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}$
$PR = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2} = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$
Since $PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = PR$,the sum of the distances between two pairs of points is equal to the distance between the third pair.
Therefore,the points $P, Q,$ and $R$ are collinear.