The equation of a line making an angle $60^{\circ}$ with the line $x+y-3=0$ and passing through the point $(1,1)$ is

Let $P$ be the pair of lines represented by $2 x^2-5 x y+2 y^2+6 x-3 y=0$ and 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 origin.

(iii) $\gamma$ is the constant term in the equation of the pair of angular bisectors of $P$.

Then,

The combined equation of the diagonals of the parallelogram formed by the lines

$$ \left(7 x^2-4 x y+8 y^2\right)^2+(4 x-8 y-32)\left(7 x^2-4 x y+8 y^2\right)=0 $$

is

If $M$ is the foot of the perpendicular drawn from the origin $O$ on to the variable line $L$, passing through a fixed point $(a, b)$, then the locus of the mid-point of $O M$ is