The equations of lines are given by  $\mathrm{sin\theta }\left(\mathrm{x}+\mathrm{a}\right)=\left(1+\mathrm{cos\theta }\right)\mathrm{y}$, $\mathrm{sin\theta }\left(\mathrm{x}-\mathrm{a}\right)=-\left(1-\mathrm{cos\theta }\right)\mathrm{y}$. The locus of point of intersection of lines is ${\mathrm{x}}^{\mathrm{m}}+{\mathrm{y}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{l}}$ then $\mathrm{l}+\mathrm{m}+\mathrm{n}$ is

Locus of point of intersection of two lines having parameter $\theta$  , is the equation obtained by eliminating $\theta$ from both equations 

$\frac{\mathrm{x}+\mathrm{a}}{\mathrm{y}}=\frac{1+\mathrm{cos\theta }}{\mathrm{sin\theta }}$
 
And 
$\frac{\mathrm{x}-\mathrm{a}}{\mathrm{y}}=-\frac{1-\mathrm{cos\theta }}{\mathrm{sin\theta }}$  
Multiply both equations
$\begin{array}{c}\frac{\mathrm{x}+\mathrm{a}}{\mathrm{y}}\cdot \frac{\mathrm{x}-\mathrm{a}}{\mathrm{y}}=-\left(\frac{1+\mathrm{cos\theta }}{\mathrm{sin\theta }}\right)\cdot \left(\frac{1-\mathrm{cos\theta }}{\mathrm{sin\theta }}\right)\\ \frac{{\mathrm{x}}^2-{\mathrm{a}}^2}{{\mathrm{y}}^2}=-\frac{1-{\cos }^2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}\\ =-1\\ {\mathrm{x}}^2-{\mathrm{a}}^2=-{\mathrm{y}}^2\\ {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2\end{array}$

Here $\mathrm{l}=\mathrm{m}=\mathrm{n}=2$ , 
Therefore, $\mathrm{l}+\mathrm{m}+\mathrm{n}=\boxed{6}$