Let the line L pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$. Then, which of the following points lies on the line $L$ ?

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 $\left.\log _{\mathrm{e}} 2=0.7\right):$