$\because$ Points $(\mathrm{k}, 1,5),(1,0,3) \&(7,-2, \mathrm{~m})$ are collinear.
$$
\begin{aligned}
& \mathrm{A} \equiv(\mathrm{k}, 1,5), \mathrm{B} \equiv(1,0,3), \mathrm{C}=(7,-2, \mathrm{~m}) \\
& \overrightarrow{\mathrm{AB}}=(1-\mathrm{k}) \hat{\mathrm{i}}+(0-1) \hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{AC}}=(7-\mathrm{k}) \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+(\mathrm{m}-5) \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{AB}}=\lambda \overrightarrow{\mathrm{AC}} \\
& \Rightarrow(1-\mathrm{k}) \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}=\lambda[(7-\mathrm{k}) \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+(\mathrm{m}-5) \hat{\mathrm{k}}]
\end{aligned}
$$
Comparing both sides, we get
$$
\begin{aligned}
& 1-\mathrm{k}=\lambda(7-\mathrm{k}) \\
& -1=-3 \lambda \Rightarrow \lambda=\frac{1}{3} \\
& -2=\lambda(\mathrm{m}-5) \Rightarrow-6=\mathrm{m}-5 \Rightarrow \mathrm{m}=-1
\end{aligned}
$$
$$
\text { from eqn. (i) } \Rightarrow 1-k=\frac{1}{3}(7-k)
$$
$$
\begin{aligned}
& \Rightarrow 3-3 \mathrm{k}=7-\mathrm{k} \Rightarrow 2 \mathrm{k}=-4 \Rightarrow \mathrm{k}=-2 \\
& \therefore(\mathrm{k}, \mathrm{m})=(-2,-1)
\end{aligned}
$$