The image of the line  $2\mathrm{x}-\mathrm{y}-1=0$ with respect to the line $3\mathrm{x}-2\mathrm{y}+4=0$  is 

The image of line $\mathrm{lx}+\mathrm{my}+\mathrm{n}=0$  with respect to the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$  is   $\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)\left(\mathrm{lx}+\mathrm{my}+\mathrm{n}\right)=2\left(\mathrm{al}+\mathrm{bm}\right)\left(\mathrm{ax}+\mathrm{by}+\mathrm{c}\right)=0$
Hence the image of the line   with respect to the line   is 
$\begin{array}{c}\left({\left(3\right)}^2+{\left(-2\right)}^2\right)\left(2\mathrm{x}-\mathrm{y}-1\right)=2\left(3\left(2\right)+\left(-2\right)\left(-1\right)\right)\left(3\mathrm{x}-2\mathrm{y}+4\right)\\ 13\left(2\mathrm{x}-\mathrm{y}-1\right)=16\left(3\mathrm{x}-2\mathrm{y}+4\right)\\ 26\mathrm{x}-13\mathrm{y}-13=48\mathrm{x}-32\mathrm{y}+64\\ 22\mathrm{x}-19\mathrm{y}+77=0\end{array}$ 
Compare $22\mathrm{x}-19\mathrm{y}+77=0$  with  $\text{px+qy+r=0}$
It implies that  $\mathrm{p}=22,\mathrm{q}=-19,\mathrm{r}=77$
Therefore,   $\mathrm{p}+\mathrm{q}+\mathrm{r}=22-19+77=\boxed{80}$