The equation to the pair of lines perpendicular to 2x2+3xy+y2=0and passing through 2,1 is x2−3xy+2y2+lx+my+n=0then l+m+n=

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