Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4 \sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance, between the lines $\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$ and $\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$, is equal to

If the domain of the function $f(x)=\sin ^{-1}\left(\frac{5-x}{3+2 x}\right)+\frac{1}{\log _e(10-x)}$ is $(-\infty, \alpha] \cup[\beta, \gamma)-\{\delta\}$, then $6(\alpha+\beta+\gamma+\delta)$ is equal to

Let $f:[1, \infty) \rightarrow \mathbb{R}$ be a differentiable function. If $6 \int\limits_1^x f(t) d t=3 x f(x)+x^3-4$ for all $x \geq 1$, then the value of $f(2)-f(3)$ is :

If the image of the point $\mathrm{P}(1,2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-\mathrm{z}}{2}$ is $\mathrm{Q}(5, b, \mathrm{c})$, then $a^2+b^2+c^2$ is equal to