If the equation of a line parallel to $3x - 2y + 5 = 0$ and at a distance of $5$ units from it is $3x - 2y + C = 0$,then $C$ is equal to

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 ..........

If $p_{1}$ and $p_{2}$ are the lengths of perpendiculars from the origin to the lines $x \sin \theta + y \cos \theta = 5 \cos 2 \theta$ and $x \operatorname{cosec} \theta + y \sec \theta - 5 = 0$ respectively,then $p_{1}^{2} + 4 p_{2}^{2} = $

$A$ straight line is parallel to the lines $3x - y - 3 = 0$ and $3x - y + 5 = 0$ and lies between them. If its distance from these lines is in the ratio $3 : 5$,then its equation is:

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ is

The length of the perpendicular from the point $(b, a)$ to the line $\frac{x}{a} - \frac{y}{b} = 1$ is: