The combined equation of the three sides of a triangle is $(x^2-y^2)(2x+3y-6)=0$. If the point $(0, \alpha)$ lies in the interior of this triangle,then

One of the lines represented by the equation ${x^2} + 6xy = 0$ is

The equation of the line common to the pair of lines $(p^2-q^2) x^2+(q^2-r^2) xy+(r^2-p^2) y^2=0$ and $(l-m) x^2+(m-n) xy+(n-l) y^2=0$ is

The equation of the pair of lines perpendicular to the lines represented by $x^2-5xy+4y^2=0$ and passing through the point $(2,1)$ is

If $\alpha x^2+2 \gamma x y+\beta y^2=0$ is the equation of a pair of lines passing through the origin and perpendicular to the pair of lines $b h x^2+a b x y+a h y^2=0$ $(a \neq 0, b \neq 0)$,then $\frac{\alpha \beta}{\gamma^2}=$

If the sum of slopes of the pair of lines given by $4x^{2} + 2hxy - 7y^{2} = 0$ is equal to the product of the slopes,then $h$ is