If the point $\left(\frac{\mathrm{k}-1}{\mathrm{k}},\frac{\mathrm{k}-2}{\mathrm{k}}\right)$ lies on the locus of z satisfying the inequality $\left|\frac{\mathrm{z}+3\mathrm{i}}{3\mathrm{z}+\mathrm{i}}\right|<1$, then the interval in which k is lies is

$\left|\frac{\mathrm{z}+3\mathrm{i}}{3\mathrm{z}+\mathrm{i}}\right|<1 \Rightarrow \left|\mathrm{x}+\mathrm{iy}+3\mathrm{i}\right| < \left|3(\mathrm{x}+\mathrm{iy})+\mathrm{i}\right|$

$\Rightarrow \left|\mathrm{x}+\mathrm{i}(\mathrm{y}+3)\right| < \left|3\mathrm{x}+\mathrm{i}(3\mathrm{y}+1)\right|$

$\Rightarrow \sqrt{{\mathrm{x}}^2+{(\mathrm{y}+3)}^2 }< \sqrt{9{\mathrm{x}}^2+{(3\mathrm{y}+1)}^2}$

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+6\mathrm{y}+9 < 9{\mathrm{x}}^2+9{\mathrm{y}}^2+6\mathrm{y}+1$

$\Rightarrow 8{\mathrm{x}}^2+8{\mathrm{y}}^2>8$

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2>1$

$\therefore {\left(\frac{\mathrm{k}-1}{\mathrm{k}}\right)}^2+{\left(\frac{\mathrm{k}-2}{\mathrm{k}}\right)}^2>1$

$\Rightarrow {\mathrm{k}}^2-2\mathrm{k}+1+{\mathrm{k}}^2-4\mathrm{k}+4>{\mathrm{k}}^2$

$\Rightarrow {\mathrm{k}}^2-6\mathrm{k}+5>0$

$\Rightarrow (\mathrm{k}-1) (\mathrm{k}-5)>0$

$$\begin{gathered} \Rightarrow \mathrm{k}<1, \mathrm{k}>5 \\ \boxed{\mathrm{k}\in (-\infty , 1)\cup (5, \infty )} \end{gathered}$$