Let the centre of the circle $x^2+y^2+2 \mathrm{~g} x+2 f y+25=0$ be in the first quadrant and lie on the line $2 x-y=4$. Let the area of an equilateral triangle inscribed in the circle be $27 \sqrt{3}$. Then the square of the length of the chord of the circle on the line $x=1$ is $\_\_\_\_$ .

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{j}-\hat{k}$ and $\vec{c}$ be three vectors such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$, then $\vec{c} \cdot(\vec{a}-2 \vec{b})$ is equal to $\_\_\_\_$ .

For the functions $f(\theta)=\alpha \tan ^2 \theta+\beta \cot ^2 \theta$, and $g(\theta)=\alpha \sin ^2 \theta+\beta \cos ^2 \theta, \alpha>\beta>0$, let $\min\limits_{0<\theta<\frac{\pi}{2}} f(\theta)=\max\limits_{0<\theta<\pi} g(\theta)$. If the first term of a G.P. is $\left(\frac{\alpha}{2 \beta}\right)$, its common ratio is $\left(\frac{2 \beta}{\alpha}\right)$ and the sum of its first 10 terms is $\frac{m}{n}, \operatorname{gcd}(m, n)=1$, then $m+n$ is equal to $\_\_\_\_$ .

Let $y=y(x)$ be the solution of the differential equation $\left(x^2-x \sqrt{x^2-1}\right) d y+\left(y\left(x-\sqrt{x^2-1}\right)-x\right) d x=0, x \geq 1$. If $y(1)=1$, then the greatest integer less than $y(\sqrt{5})$ is $\_\_\_\_$ .