The image of the line  $\mathrm{x}-\mathrm{y}+3=0$ on  x-axis is  $\mathrm{px}+\mathrm{qy}+\mathrm{r}=0$ then  the value of $\mathrm{p}+\mathrm{q}+\mathrm{r}=$

To get the image of any line ${\mathrm{L}}_1=0$  with respect to the other line  ${\mathrm{L}}_2=0$  then find the equation of the line passing through images of any two points on the line  ${\mathrm{L}}_1=0$ with respect to the line  ${\mathrm{L}}_2=0$
Two points on the line  $\mathrm{x}-\mathrm{y}+3=0$ are $\left(0,3\right)$  and  $\left(1,4\right)$
The image of a point  $\left(\mathrm{a},\mathrm{b}\right)$ on   x-axis is  $\left(\mathrm{a},-\mathrm{b}\right)$
Hence the images of   $\left(0,3\right)$ and  $\left(1,4\right)$ on   x-axis are  $\left(0,-3\right)$ and  $\left(1,-4\right)$
Equation of the line passing through the points  $\left(0,-3\right)$  and  $\left(1,-4\right)$ is
 $\begin{array}{c}\mathrm{y}+3=\frac{-4+3}{1-0}\left(\mathrm{x}-0\right)\\ \mathrm{y}+3=-\mathrm{x}\\ \mathrm{x}+\mathrm{y}+3=0\end{array}$ 
Compare this with  $\mathrm{px}+\mathrm{qy}+\mathrm{r}=0$, 
Hence $\mathrm{p}=1,\mathrm{q}=1,\mathrm{r}=3$.
Therefore $\mathrm{p}+\mathrm{q}+\mathrm{r}=1+1+3=\boxed{5}$