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

Solution:

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.