Prove that the points (0, – 1, – 7), (2, 1, – 9)

Given that three points $A(0,-1,-7), B(2,1,-9)$ and $C(6,5,-13)$ are collinear

Therefore we can write as

$\mathrm{AB}=\sqrt{(2-0)^{2}+(1-(-1))^{2}+((-9)-(-7))^{2}}=\sqrt{4+4+4}=2 \sqrt{3}$

$\mathrm{BC}=\sqrt{(6-2)^{2}+(5-1)^{2}+((-13)-(-9))^{2}}=\sqrt{16+16+16}=4 \sqrt{3}$

$\mathrm{AC}=\sqrt{(6-0)^{2}+(5-(-1))^{2}+((-13)-(-7))^{2}}=\sqrt{36+36+36}=6 \sqrt{3}$

$\Rightarrow \mathrm{AB}+\mathrm{BC}=\mathrm{AC}$

Since points A, B and C are collinear.

$A B: A C=2 \sqrt{3}: 6 \sqrt{3}=1: 3$

Hence from the lengths of $A B, B C$ and $A C$ we can say that the first point divides the join of the other two in the ratio $1: 3$ externally.