Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.

Toolbox:
• Two or more vectors are said to be collinear if their magnitudes are equal or proportional.
Step 1:
Let the position vectors of points $A,B$ and $C$ be
$\overrightarrow{OA}=\hat i+2\hat j+7\hat k$
$\overrightarrow{OB}=2\hat i+6\hat j+3\hat k$
$\overrightarrow{OC}=3\hat i+10\hat j-\hat k$
Step 2:
Now $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\qquad\quad\;\;=(2\hat i+6\hat j+3\hat k)-(\hat i+2\hat j+7\hat k)$
$\qquad\quad\;\;=\hat i+4\hat j-4\hat k$
Step 3:
$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$
$\quad\;\;\;=(3\hat i+10\hat j-\hat k)-(2\hat i+6\hat j+3\hat k)$
$\quad\;\;\;=\hat i+4\hat j-4\hat k$
Step 4:
$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}$
$\quad\;\;\;=(3\hat i+10\hat j-\hat k)-(\hat i+2\hat j+7\hat k)$
$\quad\;\;\;=2\hat i+8\hat j-8\hat k$
$\quad\;\;\;=2(\hat i+4\hat j-4\hat k)$
Step 5:
$|\overrightarrow{AB}|=\sqrt{1^2+4^2+(-4)^2}$
$\qquad=\sqrt{1+16+16}$
$\qquad=\sqrt{33}$
Step 6:
$|\overrightarrow{BC}|=\sqrt{1^2+4^2+(-4)^2}$
$\qquad=\sqrt{1+16+16}$
$\qquad=\sqrt{33}$
Step 7:
$|\overrightarrow{AC}|=\sqrt{2^2+8^2+(-8)^2}$
$\qquad=\sqrt{4+64+64}$
$\qquad=\sqrt{132}$
$\qquad=2\sqrt{33}$
Step 8:
$AB=\sqrt{33},BC=\sqrt{33},AC=2\sqrt{33}$
Thus $AB+BC=AC$
Hence A,B,C are collinear.