The distance between the lines $3x - 2y = 1$ and $6x - 4y + 9 = 0$ is

If $\alpha = 1 + \sum_{r=1}^6 (-3)^{r-1} \binom{12}{2r-1}$,then the distance of the point $(12, \sqrt{3})$ from the line $\alpha x - \sqrt{3} y + 1 = 0$ is ..........

$P$ is a point on $x+y+5=0$,whose perpendicular distance from $2x+3y+3=0$ is $\sqrt{13}$. Then the coordinates of $P$ are:

If the perpendicular distances from the points $(2, 3)$,$(4, a)$ and $(\alpha, \beta)$ to the line $3x + 4y - 3 = 0$ are equal and $4\alpha - 3\beta + 1 = 0$,then the sum of all possible values of $a$,$\alpha$,and $\beta$ is:

Given the lines $y + 2x = 3$ and $y + 2x = 5$ cut the axes at $A, B$ and $C, D$ respectively.
Statement-$1$ : $ABDC$ forms a quadrilateral and the point $(2, 3)$ lies inside the quadrilateral.
Statement-$2$ : The point $(2, 3)$ lies between the two parallel lines.

If two sides of a square are $4x + 3y - 20 = 0$ and $4x + 3y + 15 = 0$,then the area of the square is