If 2x²-3xy+y²= represents two sides of a triangle and lx+my+n=0 is the third side then locus of incentre of the triangle is

Given
3x² -5xy+py²=0 and 6x² -xy-5y²=0
Here ${\mathrm{a}}_1=3,2{\mathrm{h}}_1=-5,{\mathrm{b}}_1=\mathrm{p}{\mathrm{a}}_2=6,2{\mathrm{h}}_2=-1,{\mathrm{b}}_2=-5$
Wkt $\left|\begin{array}{cc}{\mathrm{a}}_1& 2{\mathrm{h}}_1\\ {\mathrm{a}}_2& 2{\mathrm{h}}_2\end{array}\right|\left|\begin{array}{cc}2{\mathrm{h}}_1& {\mathrm{b}}_1\\ 2{\mathrm{h}}_2& {\mathrm{b}}_2\end{array}\right|={\left|\begin{array}{cc}{\mathrm{a}}_1& {\mathrm{b}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2\end{array}\right|}^2$
On solving we get p=2, p=-25/4
2x -3xy+y =0
a=2,2h=-3,b=1
Wkt h(x²-y²)=(a-b)xy
-3x²+3xy=2xy
-3x²+3y²+2xy=0
3x²+2xy-3y²=0
By optional verification