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

Let 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 a line.

$\mathrm{PQ}=\sqrt{(1+2)^{2}+(2-3)^{2}+(3-5)^{2}}$

$=\sqrt{(3)^{2}+(-1)^{2}+(-2)^{2}}$

$=\sqrt{9+I+4}$

$=\sqrt{14}$

$\mathrm{QR}=\sqrt{(7-1)^{2}+(0-2)^{2}+(-1-3)^{2}}$

$=\sqrt{(6)^{2}+(-2)^{2}+(-4)^{2}}$

$=\sqrt{36+4+16}$

$=\sqrt{56}$

$=2 \sqrt{14}$

$\mathrm{PR}=\sqrt{(7+2)^{2}+(0-3)^{2}+(-1-5)^{2}}$

$=\sqrt{(9)^{2}+(-3)^{2}+(-6)^{2}}$

$=\sqrt{81+9+36}$

$=\sqrt{126}$

$=3 \sqrt{14}$

Here, $\mathrm{PQ}+\mathrm{QR}=\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}=\mathrm{PR}$

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