$a, b$, are the roots of $x^2-3 x+p=0$
$\therefore \quad a+b=3, a b=p \quad \ldots(i)$
$\because \quad$ Sum of the roots $=-\frac{b}{a}$,
Product of roots $=\frac{c}{a}$ in equation
$a x^2+b x+c=0$
$c, d$ are the roots of $x^2-12 x+q=0$
$c+d=12$ and $c d=q \quad \ldots(ii)$
Also $a, b, c, d$ are in G.P. Let $r$ be its common ratio $\therefore \quad b=a r, c=a r^2, d=a r^3 \quad \ldots(iii)$
Now, $\quad a+b=a+a r=3$
$c+d=a r^2+a r^3=12$
On dividing, we get
$\frac{a(1+r)}{a r^2(1+r)}=\frac{3}{12}$
$\frac{1}{r^2}=\frac{1}{4}$ or $r=2$
Now $\quad \frac{q+p}{q-p}=\frac{c d+a b}{c d-a b}$ from (i) and (ii)
Putting the values of $b, c, d$ from (iii)
$\begin{aligned}
&\frac{q+p}{q-p}=\frac{a r^2 \cdot a r^3+a \cdot a r}{a r^2 \cdot a r^3-a \cdot a r} \\
&=\frac{a^2 r\left(r^4+1\right)}{a^2 r\left(r^4-1\right)}=\frac{r^4+1}{r^4-1}=\frac{2^4+1}{2^4-1}=\frac{16+1}{16-1}=\frac{17}{15}
\end{aligned}$
$[\because r=2]$
Hence $\quad \frac{q+p}{q-p}=\frac{17}{15}$