$a x+2 y=k$ is perpendicular to $2 x-3 y+7=0$
$$
\begin{aligned}
& \therefore m_1 \cdot m_2=-1 \\
& \frac{-a}{2} \times \frac{2}{3}=-1 \Rightarrow a=3
\end{aligned}
$$
$\therefore 3 x+2 y=k$ forms triangle of area 3 sq. units
$$
3 x+2 y=k
$$
$$
\frac{x}{\left(\frac{k}{3}\right)}+\frac{y}{\left(\frac{k}{2}\right)}=1
$$
$\operatorname{Ar}(\triangle \mathrm{OAB})=\left|\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\right|=3$
$$
\begin{aligned}
& \therefore|\mathrm{OA} \times \mathrm{OB}|=6 \\
& \left|\frac{k}{3} \times \frac{k}{2}\right|=6 \Rightarrow k= \pm 6
\end{aligned}
$$

Product of values $=6 \times(-6)=-36$