Given equation $a x^2+6 x y-2 y^2=0$ represents pair of perpendicular lines and $9 x^2+2 h x y+4 y^2=0$ as pair of coincident lines.
According to question
coefficient of $\left(\mathrm{x}^2\right)+$ coefficient of $\left(\mathrm{y}^2\right)=0$ for perpendicular lines.
Then, $\mathrm{a}+(-2)=0$
$$
\Rightarrow \mathrm{a}=2 \text {. }
$$
For coincident lines, the required condition is $\mathrm{h}^2-\mathrm{ab}=0$.
Here, $a=9, b=4$ for equation $9 x^2+2 h x y+4 x^2=0$
$$
\begin{aligned}
& \mathrm{h}^2-9 \times 4=0 \\
& h^2=36 \\
& \mathrm{~h}= \pm 6
\end{aligned}
$$
for positive $\mathrm{h}=6$
$\mathrm{h}=2 \times 3=3 \mathrm{a}\{$ for $\mathrm{a}=2\}$
so, option (a) is correct option.