If the pair of straight lines given by $Ax^2+2Hxy+By^2=0$ $(H^2>AB)$ forms an equilateral triangle with the line $ax+by+c=0$,then $(A+3B)(3A+B)$ is equal to:

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,

If the angle between the pair of lines $x^2+2 \sqrt{2} x y+k y^2=0, k>0$ is $45^{\circ}$,then the area (in square units) of the triangle formed by the pair of bisectors of angles between the given lines and the line $x+2 y+1=0$ is

The equation of a line which makes an angle of $45^{\circ}$ with each of the pair of lines $xy-x-y+1=0$ is

If one of the lines given by the equation $2x^2 + axy + 3y^2 = 0$ coincides with one of those given by the equation $2x^2 + bxy - 3y^2 = 0$,while the other two lines are perpendicular to each other,then the values of $a$ and $b$ are

If the pair of lines represented by $3x^2 - 5xy + Py^2 = 0$ and $6x^2 - xy - 5y^2 = 0$ have one line in common,then the sum of all possible values of $P$ is