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 $\mathrm{l}+\mathrm{m}+\mathrm{n}=$

detailed solution

Correct option is D

The combined equation of pair of lines which are perpendicular to the pair of lines ax2+2hxy+by2=0and passing through the point x1,y1is bx−x12−2hx−x1y−y1+ay−y12=0Hence the equation of pair of lines perpendicular to the pair of lines 2x2+3xy+y2=0and passing through the point 2,1is x−22−3x−2y−1+2y−12=0Simplifying the equation x−22−3x−2y−1+2y−12=0x2−4x+4−3xy+3x+6y−6+2y2−4y+2=0x2−3xy+2y2−x+2y=0Comparing the above equation with x2−3xy+2y2+lx+my+n=0We get l=−1,m=2,n=0Therefore, l+m+n=1