Consider the lines $L_1\equiv 4x+5y-6=0, L_2\equiv 2x+3y-4=0, L_3\equiv 3x-y+2=0.$ If the line $L_1=0$ intersects the lines $L_2=0$ and $L_3=0$ at the points $A$ and $B$ respectively, then the combined equation of $OA$ and $OB$ is

We have straight lines$L_1\equiv 4x+5y-6=0 ...\left(1\right), L_2\equiv 2x+3y-4=0 ....\left(2\right)$

$L_3\equiv 3x-y+2=0 ...\left(3\right)$

As given, the line $L_1=0$ intersects the lines $L_2=0$ and $L_3=0$ at the points $A$ and $B$ respectively,

Solving $\left(1\right) \mathrm{and} \left(2\right)$

we get point $A\left(-1, 2\right)$

Solving $\left(1\right)$ and $\left(3\right)$

we get point $B\left(\frac{-4}{19}, \frac{26}{19}\right)$

Hence, the equation of line $OA$:

Passing through origin and point $A$.

$\frac{y-0}{x-0}=\frac{2-0}{-1-0} \Rightarrow 2x+y=0$

Similarly, the equation of line $OB$

$\frac{y-0}{x-0}=\frac{-{\displaystyle \frac{4}{19}}-0}{{\displaystyle \frac{26}{19}}-0} \Rightarrow 13x+2y=0$

So, the equation of pair of straight line passing through $A$ and $B$ whose point of intersection is origin, is

$\left(2x+y\right)\left(13x+2y\right)=0\Rightarrow 26x^2+17xy+2y^2=0$