Solving the given pair of straaight line we get,

$6x^2-xy-12y^2=0$

$\Rightarrow 6x^2-9xy+8xy-12y^2=0$

$\Rightarrow 3x\left(2x-3y\right)+4y\left(2x-3y\right)=0$

$\Rightarrow \left(3x+4y\right)\left(2x-3y\right)=0 ...............\left(1\right)$

Now solving the other pair of straight line we get,

$15x^2+14xy-8y^2=0$

$\Rightarrow 15x^2+20xy-6xy-8y^2=0$

$\Rightarrow 5x\left(3x+4y\right)-2y\left(3x+4y\right)=0$

$\Rightarrow \left(5x-2y\right)\left(3x+4y\right)=0 .............\left(2\right)$

Now from equation $\left(1\right) \& \left(2\right)$ we can see $3x+4y=0$ is common to both equation,

Now the combined equation the other two lines is $\left(5x-2y\right)\left(2x-3y\right)=0$

$\Rightarrow 10x^2-19xy+6y^2=0$