$$
\begin{aligned}
&x+2 y=3 \\
&\Rightarrow x+2 y-3=0 \ldots \text { (i) }
\end{aligned}
$$
And, \(5 x+k y+7=0 \ldots\) (ii)
These equations are of the following form:
$$
a 1 x+b 1 y+c 1=0, a 2 x+b 2 y+c 2=0
$$
where, \(a 1=1, b 1=2, c 1=-3\) and \(a 2=5, b 2=k, c 2=7\)
(i) For a unique solution, we must have:
$$
\therefore \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \text { i.e., } \frac{1}{5} \neq \frac{2}{k} \Rightarrow \mathrm{k} \neq 10
$$
Thus for all real values of \(k\) other than 10 , the given system of equations will have a unique solution.
(ii) In order that the given system of equations has no solution, we must have:
$$
\begin{aligned}
&\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\
&\Rightarrow \frac{1}{5} \neq \frac{2}{k} \neq \frac{-3}{7} \\
&\Rightarrow \frac{1}{5} \neq \frac{2}{k} \text { and } \frac{2}{k} \neq \frac{-3}{7} \\
&\Rightarrow \mathrm{k}=10, \mathrm{k} \neq \frac{14}{-3}
\end{aligned}
$$
Hence, the required value of \(k\) is 10 .
There is no value of \(k\) for which the given system of equations has an infinite number of solutions.
