Let $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\mathrm{L}_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $\mathrm{L}_1$ and $\mathrm{L}_2$ ?

Consider the line $$\mathrm{L}$$ passing through the points $$(1,2,3)$$ and $$(2,3,5)$$. The distance of the point $$\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$$ from the line $$\mathrm{L}$$ along the line $$\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$$ is equal to

The shortest distance between the lines $$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$$ and $$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$$ is:

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