Area of the triangle formed by the lines $3x^2-4xy+y^2=0$ and $2x-y=6$ is

If the product of the lengths of the perpendiculars drawn from the point $(-1, 5)$ to the pair of lines $2x^2 - xy + ky^2 + 6x + y + 4 = 0$ is $\frac{65}{\sqrt{26}}$,then $37k^2 + 92k$ is equal to

From the point $(3,-4)$,perpendicular lines $L_1$ and $L_2$ are drawn to each of the lines represented by $S \equiv 2x^2+3xy-2y^2-7x+y+3=0$. The area of the quadrilateral formed by the pair of lines $S=0$,$L_1$,and $L_2$ is (in square units):

If $\left(\frac{2}{3}, 0\right)$ is the centroid of the triangle formed by the lines $4x^2-y^2=0$ and $lx+my+n=0$,then $l+m+n=$

The lines $(lx + my)^2 - 3(mx - ly)^2 = 0$ and $lx + my + n = 0$ form

The combined equation of the diagonals of the parallelogram formed by the lines $(7 x^2-4 x y+8 y^2)^2+(4 x-8 y-32)(7 x^2-4 x y+8 y^2)=0$ is