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).

$\therefore \overrightarrow{\mathrm{AB}}=(2-1) \hat{i}+(6-2) \hat{j}+(3-7) \hat{k}=\hat{i}+4 \hat{j}-4 \hat{k}$

$\overrightarrow{\mathrm{BC}}=(3-2) \hat{i}+(10-6) \hat{j}+(-1-3) \hat{k}=\hat{i}+4 \hat{j}-4 \hat{k}$

$\overrightarrow{\mathrm{AC}}=(3-1) \hat{i}+(10-2) \hat{j}+(-1-7) \hat{k}=2 \hat{i}+8 \hat{j}-8 \hat{k}$

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

$|\overrightarrow{\mathrm{BC}}|=\sqrt{1^{2}+4^{2}+(-4)^{2}}=\sqrt{1+16+16}=\sqrt{33}$

$|\overrightarrow{\mathrm{AC}}|=\sqrt{2^{2}+8^{2}+8^{2}}=\sqrt{4+64+64}=\sqrt{132}=2 \sqrt{33}$

$\therefore|\overrightarrow{\mathrm{AC}}|=|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|$

Hence, the given points A, B, and C are collinear.