If the area of the region bounded by the curves $y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$ is equal to $\alpha$, then $6 \alpha$. equals

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$, then the sum of all possible values of $\alpha$ is

Let $x=-1$ and $x=2$ be the critical points of the function $f(x)=x^3+a x^2+b \log _{\mathrm{e}}|x|+1, x \neq 0$. Let $m$ and M respectively be the absolute minimum and the absolute maximum values of $f$ in the interval $\left[-2,-\frac{1}{2}\right]$. Then $|\mathrm{M}+m|$ is equal to $\left(\right.$ Take $\left.\log _{\mathrm{e}} 2=0.7\right):$

Let the angle $\theta, 0<\theta<\frac{\pi}{2}$ between two unit vectors $\hat{a}$ and $\hat{b}$ be $\sin ^{-1}\left(\frac{\sqrt{65}}{9}\right)$. If the vector $\vec{c}=3 \hat{a}+6 \hat{b}+9(\hat{a} \times \hat{b})$, then the value of $9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})$ is