Solution:

Given equations are
$\begin{array}{ll}
2 \mathrm{x}+3 \mathrm{y}-7=0\ldots \ldots(\mathrm{i}) \\
(\mathrm{k}-1) \mathrm{x}+(\mathrm{k}+2) \mathrm{y}-3 \mathrm{k}=0 & \ldots \ldots(\mathrm{ii}) \\
\end{array}$
Which is of the form $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$, where $a_{1}-2, b_{1}-3, c_{1}–7, a_{2}-k-1, b_{2}-k+2$ and $c_{2}–3 k$
For the pair of linear equations to have infinitely many solutions, we must have
$\begin{array}{l}
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\
\Rightarrow \frac{2}{k-1}=\frac{3}{k+2}=\frac{-7}{-3 k} \\
\Rightarrow \frac{2}{k-1}=\frac{3}{k+2}, \frac{3}{k+2}=\frac{-7}{-3 k} \text { and } \frac{2}{k-1}=\frac{-7}{-3 k} \\
\Rightarrow 2(\mathrm{k}+2)=3(\mathrm{k}-1), 9 \mathrm{k}=7 \mathrm{k}+14 \text { and } 6 \mathrm{k}=7 \mathrm{k}-7 \\
\Rightarrow \mathrm{k}=7, \mathrm{k}=7 \text { and } \mathrm{k}=7
\end{array}$
As a result, $\mathrm{k}=7$