The equation of the straight line passing through the point $(3,2)$ and inclined at an angle of $60^{\circ}$ with the line $\sqrt{3} x+y=1$ is

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

If $\theta$ is the acute angle between the lines joining the origin to the points of intersection of the curve $x^2+x y+y^2+x+3 y+1=0$ and the straight line $x+y+2=0$, then $\cos \theta=$