By shifting the origin to the point $(2,3)$ through translation of axes. If the equation or the curve $x^2+3 x y-2 y^2+4 x-y-20=0$ is transformed to the form $A x^2+B x y+C y^2+D x+E y+F=0$, then $D+E+F=$

The points $(2,3)$ and $\left(-4,-\frac{4}{3}\right)$ lie on the opposite sides of the line $L \equiv 5 x-6 y+k=0$ and k is an integer. If the points $(1,2)$ and $(4,5)$ lie on the same side of the line $L=0$, then the perpendicular distance from origin to the line $L=0$ is

If the incentre of the triangle formed by the lines $x-2=0, x+y-1=0, x-y+3=0$ is $(\alpha, \beta)$, then $\beta=$

If the equation of the pair of straight lines intersecting at ( $a, b$ ) and perpendicular to the pair of lines $3 x^2-4 x y+5 y^2=0$ is $l x^2+2 n x y+m y^2-32 x-26 y+c=0$, then $\frac{a+b+c}{l+h+m}=$