If $\alpha, \beta(\alpha>\beta)$ are two values of $k$ such that the equations $2 x+(3-2 k) y+(2 k+1)=0$ and $k x+(k-1) y-4=0$ represents two perpendicular lines, then $\alpha^2+2 \beta=$

If $k=\frac{a+b}{a b}$ is a non-zero constant, then the point which lies on the straight line $\frac{x}{a}+\frac{y}{b}=1$ is

The point of concurrence of all the chords of the curve $3 x^2-y^2-2 x+4 y=0$ which subtend a right angle at the origin is

Let $d$ be the distance between the parallel lines $3 x-2 y+5=0$ and $3 x-2 y+5+2 \sqrt{13}=0$.

Let $L_1 \equiv 3 x-2 y+k_1=0\left(k_1>0\right)$ and $L_2 \equiv 3 x-2 y+k_2=0\left(k_2>0\right)$ be two lines that are at the distance of $\frac{4 d}{\sqrt{13}}$ and $\frac{3 d}{\sqrt{13}}$ from the line $3 x-2 y+5=0$.

Then, the combined equation of the lines $L_1=0$ and $L_2=0$ is