The position vectors of points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are
$$
\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}, 2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, 3 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-\hat{\mathrm{k}}
$$
$$
\begin{aligned}
&\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \\
&\overline{\mathrm{BC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OB}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \\
&\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OB}}=2(\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}) \\
&|\overline{\mathrm{AB}}|=\sqrt{1^2+4^2+(-4)^2}=\sqrt{1+16+16}=\sqrt{33}, \\
&|\overline{\mathrm{BC}}|=\sqrt{1^2+4^2+(-4)^2}=\sqrt{1+16+16}=\sqrt{33} \\
&|\overline{\mathrm{AC}}|=\sqrt{2^2+8^2+(-8)^2}=\sqrt{4+64+64}=2 \sqrt{33}
\end{aligned}
$$
Thus $\mathrm{AB}+\mathrm{BC}=\mathrm{AC}$. Hence $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are colinear.