Question
Show that the three points $\mathrm{A}(1,-2,3), \mathrm{B}(2,3,-4)$ and $\mathrm{C}(0,-7,10)$ are collinear.

Answer

If $\bar{a}, \bar{b}$ and $\bar{c}$ are the position vectors of the points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively, then
$
\begin{aligned}
\bar{a} & =\hat{i}-2 \hat{j}+3 \hat{k} \\
\bar{b} & =2 \hat{i}+3 \hat{j}-4 \hat{k} \\
\bar{c} & =0 \hat{i}-7 \hat{j}+10 \hat{k} \\
\overline{\mathrm{AB}} & =\bar{b}-\bar{a}=\hat{i}+5 \hat{j}-7 \hat{k} \\
\overline{\mathrm{AC}} & =\bar{c}-\bar{a} \\
& =-\hat{i}-5 \hat{j}+7 \hat{k} \\
& =(-1)[\hat{i}+5 \hat{j}-7 \hat{k}] \\
\overline{\mathrm{AC}} & =(-1) \overline{\mathrm{AB}}
\end{aligned}
$
... from(1)
That is, $\overline{\mathrm{AC}}$ is a scalar multiple of $\overline{\mathrm{AB}}$. Therefore, they are parallel. But point $\mathrm{A}$ is in common. Hence, the points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear.

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