Let the points are $\mathrm{A}(\mathrm{k},-10,3), \mathrm{B}(1,-1,3)$ and $\mathrm{C}(3,5,3)$.
So, $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}$
$$
\begin{aligned}
&=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(\mathrm{k} \hat{\mathrm{i}}-10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\
&=(1-\mathrm{k}) \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \\
&\text { and }|\overrightarrow{\mathrm{AB}}|=\sqrt{(1-\mathrm{k})^2+(9)^2+0} \\
&=\sqrt{(1-\mathrm{k})^2+81} \\
&\overline{\mathrm{BC}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \\
&\therefore|\overrightarrow{\mathrm{BC}}|=\sqrt{2^2+6^2+0}=2 \sqrt{10} \\
&\overline{\mathrm{AC}}=(3-\mathrm{k}) \hat{\mathrm{i}}+15 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \\
&\therefore|\overrightarrow{\mathrm{AC}}|=\sqrt{(3-\mathrm{k})^2+225} \\
&\text { For }|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|=|\overrightarrow{\mathrm{AC}}|,[\because \mathrm{A}, \mathrm{B}, \text { C are collinear }] \\
&\sqrt{(1-\mathrm{k})^2+81}+2 \sqrt{10}=\sqrt{(3-\mathrm{k})^2+225}
\end{aligned}
$$
$\Rightarrow \quad \sqrt{(3-\mathrm{k})^2+225}-\sqrt{(1-\mathrm{k})^2+81}=2 \sqrt{10}$
$\Rightarrow \quad \sqrt{9+\mathrm{k}^2-6 \mathrm{k}+225}-\sqrt{1+\mathrm{k}^2-2 \mathrm{k}+81}=2 \sqrt{10}$
$\Rightarrow \quad-4 \mathrm{k}+192=4 \sqrt{10} \sqrt{\mathrm{k}^2+234-6 \mathrm{k}}$
$\Rightarrow \quad-\mathrm{k}+48=\sqrt{10} \sqrt{\mathrm{k}^2+234-6 \mathrm{k}}$

$$
\begin{aligned}
&\text { Squaring both sides, we get }\\
&48 \times 48+\mathrm{k}^2-96 \mathrm{k}=10\left(\mathrm{k}^2+234-6 \mathrm{k}\right)\\
&\Rightarrow-9 \mathrm{k}^2-36 \mathrm{k}=-48 \times 48+2340\\
&\mathrm{k}^2+4 \mathrm{k}+4=0\\
&\Rightarrow(\mathrm{k}+2)^2=0\\
&\therefore \quad \mathrm{k}=-2
\end{aligned}
$$