Let $A(1,0), B(2,-1)$ and $C\left(\frac{7}{3}, \frac{4}{3}\right)$ be three points. If the equation of the bisector of the angle ABC is $\alpha x+\beta y=5$, then the value of $\alpha^2+\beta^2$ is

Let $\mathrm{A}(1,2)$ and $\mathrm{C}(-3,-6)$ be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line $7 x-y=14$. If $\mathrm{B}(\alpha, \beta)$ and $\mathrm{D}(\gamma, \delta)$ are the other two vertices, then $|\alpha+\beta+\gamma+\delta|$ is equal to :

A rectangle is formed by the lines $x=0, y=0, x=3$ and $y=4$. Let the line L be perpendicular to $3 x+y+6=0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $\left(\frac{1}{2},-5\right)$ from the line $L$ is equal to :

Among the statements

$(S 1)$ : If $A(5,-1)$ and $B(-2,3)$ are two vertices of a triangle, whose orthocentre is $(0,0)$, then its third vertex is $(-4,-7)$

and

(S2) : If positive numbers $2 a, b, c$ are three consecutive terms of an A.P., then the lines $a x+b y+c=0$ are concurrent at $(2,-2)$,