Show that the following points are collinear :

Solution:

To prove: the 3 points are collinear.

Formula: The distance between two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{Z}_{1}\right)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ is given by

$\mathrm{D}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

Here,

$\left(x_{1}, y_{1}, z_{1}\right)=(-2,3,5)$

$\left(x_{2}, y_{2}, z_{2}\right)=(1,2,3)$

$\left(x_{3}, y_{3}, z_{3}\right)=(7,0,-1)$

Length $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

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

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

$=\sqrt{9+1+4}$

$=\sqrt{14}$

Length $\mathrm{BC}=\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}+\left(z_{3}-z_{2}\right)^{2}}$

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

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

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

$=\sqrt{56}=2 \sqrt{14}$

Length $\mathrm{AC}=\sqrt{\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}+\left(z_{3}-z_{1}\right)^{2}}$

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

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

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

$=\sqrt{126}=3 \sqrt{14}$

$A B+B C=\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}=A C$

Therefore A, B, C are collinear.