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

detailed solution

Correct option is B

The combined equation of pair of lines which are perpendicular to the pair of lines ax2+2hxy+by2=0and passing through the point x1,y1 isbx−x12−2hx−x1y−y1+ay−y12=0Hence the equation of pair of lines perpendicular to the pair of lines x2+xy+y2=0 and passing through the point (2,1), is x−22−x−2y−1+y−12=0Simplifying the equation x2−4x+4−xy+x+2y−2+y2−2y+1=0x2−xy+y2−3x+3=0Comparing the above equation with x2−xy+y2+lx+my+n=0We get l=−3,m=0,n=3Therefore, l+m+n=0