$4 p^{2}+4 p q-3 q^{2}-36=0$
$\Rightarrow\left(4 p^{2}+4 p q+q^{2}\right)-4 q^{2}-36=0$
$(2 p+q)^{2}=4 q^{2}+36$
$2 p+3 q=18$
$p=\frac{18-3 q}{2}$
$\Rightarrow\left[\frac{18-3 \mathrm{q}}{2} \times 2+\mathrm{q}\right]^{2}=4 \mathrm{q}^{2}+36$
$\Rightarrow(18-3 \mathrm{q}+\mathrm{q})^{2}$
$\begin{aligned} &(18-2 q)^{2}=4 q^{2}+36 \\ & 324+4 q^{2}-72 q=4 q^{2}+36 \\ 72 q=& 324-36=288 \end{aligned}$
$\mathrm{q}=\frac{288}{72}=4$
Putting the value of $\mathrm{q}$ in $2 \mathrm{p}+3 \mathrm{q}=18$ $2 p+3 \times 4=18$
$2 p=18-12$
$\mathrm{p}=\frac{6}{2}=3$
$(2 p+q)=2 \times 3+4=10$
Option (c) is correct