The equation of angular bisector of the lines $3\mathrm{x}+4\mathrm{y}+5=0$  and $4\mathrm{x}-3\mathrm{y}+7=0$  is $\mathrm{lx}+\mathrm{my}+\mathrm{n}=0$ , which makes acute angle with  x-axis then the value of $\mathrm{l}+\mathrm{m}+\mathrm{n}$  is 

The equations of angular bisectors of  ${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0$ and  ${\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0$ are   $\frac{{\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1}{\sqrt{{{\mathrm{a}}_1}^2+{{\mathrm{b}}_1}^2}}=\pm \frac{{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2}{\sqrt{{{\mathrm{a}}_2}^2+{{\mathrm{b}}_2}^2}}$
To get the equation of angular bisectors of two perpendicular lines, add both equations for one angular bisector and subtract one from the other for other angular bisector.

Since the lines given are perpendicular to each other

The angular bisector is

 $\begin{array}{c}3\mathrm{x}+4\mathrm{y}+5+4\mathrm{x}-3\mathrm{y}+7=0\\ 7\mathrm{x}+\mathrm{y}+12=0\end{array}$

The other angular bisector is 
$\begin{array}{c}3\mathrm{x}+4\mathrm{y}+5-\left(4\mathrm{x}-3\mathrm{y}+7\right)=0\\ 3\mathrm{x}+4\mathrm{y}+5-4\mathrm{x}+3\mathrm{y}-7=0\\ -\mathrm{x}+7\mathrm{y}-2=0\\ \mathrm{x}-7\mathrm{y}+2=0\end{array}$

Observing the above two angular bisectors, the slope of  $\mathrm{x}-7\mathrm{y}+2=0$ is positive, hence this makes acute angle with   x-axis. 
Therefore,   $\mathrm{l}+\mathrm{m}+\mathrm{n}=1-7+2=\boxed{-4}$