Find the coordinates of the point where the line $3x - 2y - 6 = 0$ intersects the $Y$-axis.

The equation of the line parallel to the line $3x - 4y + 2 = 0$ and passing through $(-2, 3)$ is

Reduce the equation $y-2=0$ into the normal form $x \cos \omega + y \sin \omega = p$. Find the perpendicular distance from the origin $(p)$ and the angle between the perpendicular and the positive $x$-axis $(\omega)$.

The length of the perpendicular from the origin to a line is $9$ and the perpendicular makes an angle of $120^{\circ}$ with the positive direction of the $y$-axis. Find the equation of the line.

$P(6,4)$ is a point on the line $x-y-2=0$. If $A(\alpha, \beta)$ and $B(\gamma, \delta)$ are two points on this line lying on either side of $P$ at a distance of $4$ units from $P$,then $\alpha^2+\beta^2+\gamma^2+\delta^2=$

Find the equation of the line passing through $(4, 6)$ and parallel to the line $3x - 7y + 2 = 0$.