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

If $O$ be the origin, then
$\overrightarrow {OA} = \hat i + 2\hat j + 7\hat k,\,\,\,\overrightarrow {OB} = 2\hat i + 6\hat j + 3\hat k$ and $\overrightarrow {OC} = 3\hat i + 10\hat j - \hat k$

$\therefore$ $\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = (2\hat i + 6\hat j + 3\hat k) - (\hat i + 2\hat j + 7\hat k) = \hat i + 4\hat j - 4\hat k$

$\overrightarrow {BC} = \overrightarrow {OC} - \overrightarrow {OB} = (3\hat i + 10\hat j - \hat k) - (2\hat i + 6j + 3\hat k) = \hat i + 4j - 4\hat k$

and
$\overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} = (3\hat i + 10\hat j - \hat k) - (\hat i + 2\hat j + 7\hat k) = 2\hat i + 8\hat j - 8\hat k$

$\therefore$ $AB = |\overrightarrow {AB} | = \sqrt {{1^2} + {4^2} + {{\left( { - 4} \right)}^2}} = \sqrt {1 + 16 + 16} = \sqrt {33}$

and $AC = |\overrightarrow {AC} | = \sqrt {{2^2} + {8^2} + {{( - 8)}^2}} = \sqrt {4 + 64 + 64}$ $= \sqrt {132} = 2\sqrt {33}$

Clearly, $AB + BC = AC$. Hence, $A,B,C$ are collinear.