An equilateral triangle OAB is inscribed in the parabola $y^2=4 x$ with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having $A B$ as a diameter from the origin is

The system of linear equations

$$ \begin{aligned} & x+y+z=6 \\ & 2 x+5 y+a z=36 \\ & x+2 y+3 z=b \end{aligned} $$

has :

If the image of the point $\mathrm{P}(a, 2, a)$ in the line $\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}$ is Q and the image

of Q in the line $\frac{x-2 b}{2}=\frac{y-a}{1}=\frac{z+2 b}{-5}$ is P , then $a+b$ is equal to $\_\_\_\_$ .

If the solution curve $y=f(x)$ of the differential equation

$$ \left(x^2-4\right) y^{\prime}-2 x y+2 x\left(4-x^2\right)^2=0, x>2, $$

passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$