Given that:
$\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3}=a+\frac{b}{x+p}+\frac{c}{q x+3}+\frac{d}{3 x+p}$ ...(i)
$=\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3}=1+\frac{6 x^2-4 x+8}{6 x^3+x^2-10 x+3}$
$=1+\frac{6 x^2-4 x+8}{(x-1)(3 x-1)(2 x+3)}$ ...(ii)
Now let $\frac{6 x^2-4 x+8}{(x-1)(3 x-1)(2 x+3)}=\frac{A}{(x-1)}+\frac{B}{(3 x-1)}+\frac{C}{(2 x+3)}$ ...(iii)
$\Rightarrow\left(6 x^2-4 x+8\right)= \mathrm{A}(3 x-1)(2 x+3)+\mathrm{B}(x-1)(2 x+3)+\mathrm{C}(x-1)(3 x-1)$ ...(iv)
Put $x=1$ in equation (iv), we get
$14-4=A(2)(5) \Rightarrow A=1$
Put $x=\frac{1}{3}$ in $\mathrm{eq}^{\mathrm{n}}$ (iv), we get
$\frac{22}{3}=\mathrm{B}\left(-\frac{22}{9}\right) \Rightarrow \mathrm{B}=-3$
Put $x=-\frac{3}{2}$ in equation (iv), we get $\frac{55}{2}=$ C. $\frac{55}{} \Rightarrow \mathrm{C}=2$
Using the value of A, B and C in equation (iii), we get
$\frac{6 x^2-4 x+8}{6 x^3+x^2-10 x+3}=\frac{1}{x-1}-\frac{3}{3 x-1}+\frac{2}{2 x+3}$ ...(iv)
From $\mathrm{eq}^{\mathrm{n}}$ (ii) & (v), we get
$\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3}=1+\frac{1}{x-1}-\frac{3}{3 x-1}+\frac{2}{2 x+3}$ ...(vi)
From $\mathrm{eq}^{\mathrm{n}}$ (i) & (vi), we get
$\begin{aligned} & a=1, b=1, c=2, d=-3, p=-1, q=2 \\ & \frac{a+b}{p+q}=\frac{1+1}{-1+2}=\frac{2}{1}=2\end{aligned}$