Let $e$ be the base of natural logarithm and let $f:\{1,2,3,4\} \rightarrow\left\{1, e, e^2, e^3\right\}$ and $\mathrm{g}:\left\{1, e, e^2, e^3\right\} \rightarrow\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $\phi(x)=\left[f^{-1}\left\{g^{-1}\left(\frac{1}{2}\right)\right\}\right]^x$, then the area of the region $\mathrm{R}=\left\{(x, y): x^2 \leq y \leq \phi(x), 0 \leq x \leq 1\right\}$ is :

Let $A=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1\end{array}\right]$ satisfy

$\mathrm{A}^2+\alpha(\operatorname{adj}(\operatorname{adj}(\mathrm{A})))+\beta(\operatorname{adj}(\mathrm{A})(\operatorname{adj}(\operatorname{adj}(\mathrm{A}))))=\left[\begin{array}{ccc}2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1\end{array}\right]$ for some $\alpha, \beta \in \mathbb{R}$.

Then $(\alpha-\beta)^2$ is equal to $\_\_\_\_$

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 $\_\_\_\_$ .