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

The given points are $A(1, 2, 7)$,$B(2, 6, 3)$,and $C(3, 10, -1)$.
First,we find the vectors $\overrightarrow{AB}$,$\overrightarrow{BC}$,and $\overrightarrow{AC}$:
$\overrightarrow{AB} = (2-1)\hat{i} + (6-2)\hat{j} + (3-7)\hat{k} = \hat{i} + 4\hat{j} - 4\hat{k}$
$\overrightarrow{BC} = (3-2)\hat{i} + (10-6)\hat{j} + (-1-3)\hat{k} = \hat{i} + 4\hat{j} - 4\hat{k}$
$\overrightarrow{AC} = (3-1)\hat{i} + (10-2)\hat{j} + (-1-7)\hat{k} = 2\hat{i} + 8\hat{j} - 8\hat{k}$
Now,calculate the magnitudes of these vectors:
$|\overrightarrow{AB}| = \sqrt{1^2 + 4^2 + (-4)^2} = \sqrt{1 + 16 + 16} = \sqrt{33}$
$|\overrightarrow{BC}| = \sqrt{1^2 + 4^2 + (-4)^2} = \sqrt{1 + 16 + 16} = \sqrt{33}$
$|\overrightarrow{AC}| = \sqrt{2^2 + 8^2 + (-8)^2} = \sqrt{4 + 64 + 64} = \sqrt{132} = 2\sqrt{33}$
Since $|\overrightarrow{AC}| = |\overrightarrow{AB}| + |\overrightarrow{BC}|$ (i.e.,$2\sqrt{33} = \sqrt{33} + \sqrt{33}$),the points $A$,$B$,and $C$ are collinear.