If the pair of straight lines 9 x 2 + a x y + 4 y 2 + 6 x + b y - 3 = 0 represents two parallel lines then

Given,

$9x^2+axy+4y^2+6x+by-3=0$ represents two parallel lines,

So, $△=0\Rightarrow 2fgh+abc-a{f}^2-b{g}^2-c{h}^2=0$ on comparing with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

We get, $△=a\times 3\times \frac{b}{2}+9\times 4\times \left(-3\right)-9{\left(\frac{b}{2}\right)}^2-4\times 9+3{\left(\frac{a}{2}\right)}^2=0$

$\Rightarrow \frac{ab}{2}-\frac{3b^2}{4}+\frac{a^2}{4}=48 ...........\left(1\right)$

Also for parallel lines $h^2=ab\Rightarrow \frac{a^2}{4}=36\Rightarrow a=\pm 12 .........\left(2\right)$

Now on solving equation $\left(1\right) \& \left(2\right)$ we get,

$a=12 \& b=4$