If the equation ${\mathrm{\lambda x}}^2-5\mathrm{xy}+6{\mathrm{y}}^2+\mathrm{x}-3\mathrm{y}=0$ represents a pair of straight lines then the point of intersection is

$\mathrm{f}(\mathrm{x},\mathrm{y})\equiv {\mathrm{\lambda x}}^2-5\mathrm{xy}+6{\mathrm{y}}^2+\mathrm{x}-3\mathrm{y}=0$ represents a pair of straight lines

$\frac{\mathrm{\partial }\mathrm{f}}{\mathrm{\partial }\mathrm{x}}=2\mathrm{\lambda x}-5\mathrm{y}+1=0$               ….. (1)

$\frac{\mathrm{\partial }\mathrm{f}}{\mathrm{\partial }\mathrm{y}}=-5\mathrm{x}+12\mathrm{y}-3=0$           …… (2)

we find $\mathrm{\lambda }$ by

$\mathrm{\Delta }\equiv \mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2=0$

(or)

$\left|\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{h}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{g}\\ \mathrm{h}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{f}\\ \mathrm{g}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{f}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{c}\end{array}\right|=0$

$\begin{array}{c}\left|\begin{array}{ccc}\mathrm{\lambda }& -\frac{5}{2}& \frac{1}{2}\\ \frac{-5}{2}& 6& \frac{-3}{2}\\ \frac{1}{2}& \frac{-3}{2}& 0\end{array}\right|=\frac{1}{2}\left[\frac{15}{4}-\frac{6}{2}\right]+\frac{3}{2}\left[\frac{-3\mathrm{\lambda }}{2}+\frac{5}{4}\right]=0\\ =\frac{3}{4}-3\left(\frac{6\mathrm{\lambda }-5}{4}\right)=0\\ \therefore \mathrm{\lambda }=1.\end{array}$

Now from (1) and (2)

$2\mathrm{x}-5\mathrm{y}+1=0$          …… (1)

$-5\mathrm{x}+12\mathrm{y}-3=0$    ……. (2)

$\frac{\mathrm{x}}{3}=\frac{\mathrm{y}}{1}=\frac{1}{-1}$

$\therefore$ Point of intersection is $(-3,-1)$