The equation to the pair of lines perpendicular to $2{\mathrm{x}}^2+3\mathrm{xy}+{\mathrm{y}}^2=0$and passing through $\left(2,1\right)$ is ${\mathrm{x}}^2-3\mathrm{xy}+2{\mathrm{y}}^2+\mathrm{lx}+\mathrm{my}+\mathrm{n}=0$then $\mathrm{l}+\mathrm{m}+\mathrm{n}=$

The combined equation of pair of lines which are perpendicular to the pair of lines ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2=0$and passing through the point $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$is 
$\mathrm{b}{\left(\mathrm{x}-{\mathrm{x}}_1\right)}^2-2\mathrm{h}\left(\mathrm{x}-{\mathrm{x}}_1\right)\left(\mathrm{y}-{\mathrm{y}}_1\right)+\mathrm{a}{\left(\mathrm{y}-{\mathrm{y}}_1\right)}^2=0$
Hence the equation of pair of lines perpendicular to the pair of lines $2{\mathrm{x}}^2+3\mathrm{xy}+{\mathrm{y}}^2=0$and passing through the point $\left(2,1\right)$is ${\left(\mathrm{x}-2\right)}^2-3\left(\mathrm{x}-2\right)\left(\mathrm{y}-1\right)+2{\left(\mathrm{y}-1\right)}^2=0$
Simplifying the equation 
$\begin{array}{c}{\left(\mathrm{x}-2\right)}^2-3\left(\mathrm{x}-2\right)\left(\mathrm{y}-1\right)+2{\left(\mathrm{y}-1\right)}^2=0\\ {\mathrm{x}}^2-4\mathrm{x}+4-3\mathrm{xy}+3\mathrm{x}+6\mathrm{y}-6+2{\mathrm{y}}^2-4\mathrm{y}+2=0\\ {\mathrm{x}}^2-3\mathrm{xy}+2{\mathrm{y}}^2-\mathrm{x}+2\mathrm{y}=0\end{array}$
Comparing the above equation with ${\mathrm{x}}^2-3\mathrm{xy}+2{\mathrm{y}}^2+\mathrm{lx}+\mathrm{my}+\mathrm{n}=0$
We get $\mathrm{l}=-1,\mathrm{m}=2,\mathrm{n}=0$
Therefore, $\mathrm{l}+\mathrm{m}+\mathrm{n}=\boxed{1}$