If angle $\theta$ in $[0,2 \pi]$ satisfies both the equations $\cot \theta=\sqrt{3}$ and $\sqrt{3} \sec \theta+2=0$, then $\theta$ is equal to

The area of the triangle with vertices $(1,2,0)$, $(1,0,2)$ and $(0,3,1)$ is

The particular solution of the differential equation, $x y \frac{\mathrm{~d} y}{\mathrm{~d} x}=x^2+2 y^2$ when $y(1)=0$ is

The equation of the tangent to the circle, given by $x=5 \cos \theta, y=5 \sin \theta$ at the point $\theta=\frac{\pi}{3}$ on it , is