Given The system of equations $x+2 y=3$,
$3 x+6 y=a-2$ has no solution.
Since, we know that simultaneous linear equations
$a_1 x+b_1 y=c_1$ and $a_2 x+b_2 y=c_2$ has no solution, if $\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
Hence, from the given equations,
$$
\begin{aligned}
\frac{1}{3} & =\frac{2}{6} \neq \frac{3}{a-2} \\
\Rightarrow \quad \frac{1}{3} & =\frac{1}{3} \neq \frac{3}{a-2}
\end{aligned}
$$

To satisfy above condition, $a-2 \neq 9$
$$
\Rightarrow \quad a \neq 11
$$