Let the origin be the centroid of an equilateral triangle $ABC$ and one of its sides be along the straight line $x+y=3$. If $R$ and $r$ are its circumradius and inradius respectively,then $R+r=$

The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is:

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ represents a line perpendicular to the line $x + y + 1 = 0$. If $p$ is positive,$\alpha$ lies in the fourth quadrant,and the perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line $L = 0$ is $5$ units,then $p =$

Let the centroid of an equilateral triangle $ABC$ be at the origin. Let one of the sides of the equilateral triangle be along the straight line $x + y = 3$. If $R$ and $r$ are the radius of the circumcircle and incircle respectively of $\Delta ABC$,then $(R + r)$ is equal to ..... .

If $p$ and $q$ are lengths of the perpendiculars from the origin to the lines $x \sec \theta + y \operatorname{cosec} \theta = k$ and $x \cos \theta - y \sin \theta = k \cos 2\theta$ respectively,then

An equilateral triangle is constructed between the lines $\sqrt{3} x+y-6=0$ and $\sqrt{3} x+y+9=0$ with the base on one line and the vertex on the other. The area (in sq. units) of the triangle so formed is