Equation of one of the line through the point $\left(1,5\right)$ and making an angle of $45°$  with the line$\mathrm{x}-3\mathrm{y}+5=0$ is $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ . If  $\mathrm{a}>0,\mathrm{b}>0$ then the value of $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is

The equation of any line which makes an angle  $\alpha$ with the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$  and passing through the point $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ is $\mathrm{y}-{\mathrm{y}}_1=\tan \left(\mathrm{\theta }\pm \mathrm{\alpha }\right)\left(\mathrm{x}-{\mathrm{x}}_1\right)$ here   
Here  $\mathrm{tan\theta }=-\frac{\mathrm{a}}{\mathrm{b}}$ and   $\alpha =45°$
Equations of the required lines are  $y-5=\tan \left(\theta \pm 45°\right)\left(x-1\right)$
It implies 
  $\begin{array}{c}y-5=\tan \left(\theta +45°\right)\left(x-1\right)\\ =\frac{1+\tan \theta }{1-\tan \theta }\left(x-1\right)\end{array}$ 
Substitute $\tan \theta =\frac{1}{3}$

$\begin{array}{c}y-5=\frac{1+\frac{1}{3}}{1-\frac{1}{3}}\left(x-1\right)\\ =2\left(x-1\right)\\ y-5=2x-2\\ 2x-y+3=0\end{array}$

For the other line equation

$\begin{array}{c}y-5=\tan \left(\theta -45°\right)\left(x-1\right)\\ =\frac{\tan \theta -1}{1+\tan \theta }\left(x-1\right)\end{array}$

Substitute $\tan \theta =\frac{1}{3}$

$\begin{array}{c}y-5=\frac{\frac{1}{3}-1}{1+\frac{1}{3}}\left(x-1\right)\\ x+2y-11=0\end{array}$

Since $a>0,b>0$, the sum of the values of $a,b,c$  is  $1+2-11=\boxed{-8}$