If the area of the triangle formed by the lines $y=x+c$ and $2x^2+5xy+3y^2=0$ is $\frac{1}{20}$ sq. units,then $c=$

The figure formed by the pairs of lines $6x^2+13xy+6y^2=0$ and $6x^2+13xy+6y^2+10x+10y+4=0$ is a

The triangle formed by $x^2-4xy+y^2=0$ and $x+y+4\sqrt{6}=0$ is

If the equation of the pair of straight lines passing through the point $(1,1)$ and perpendicular to the pair of lines $3x^2+11xy-4y^2=0$ is $ax^2+2hxy+by^2+2gx+2fy+12=0$,then $2(a-h+b-g+f-12)=$

Let $P$ be the pair of lines represented by $2x^2 - 5xy + 2y^2 + 6x - 3y = 0$. Consider the following independent statements:
$(i)$ $\alpha$ is the $x$-coordinate of the point of intersection of the pair of lines $P$.
(ii) $\beta$ is the slope of one of the lines of $P$ passing through the origin.
(iii) $\gamma$ is the constant term in the equation of the pair of angular bisectors of $P$.
Then,

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a