The shortest distance between the lines $r = (3t - 4)\hat{i} - 2\hat{j} - (1 + 2t)\hat{k}$ and $r = (6 + s)\hat{i} + (2 - 2s)\hat{j} + 2(1 + s)\hat{k}$ is

The distance of the point $(1, 2, -4)$ from the line $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6}$ is

The point of intersection of the lines $\vec{r}=2 \vec{b}+t(6 \vec{c}-\vec{a})$ and $\vec{r}=\vec{a}+s(\vec{b}-3 \vec{c})$ is

The shortest distance between the lines $\overline{r}=(3 \bar{i}-5 \bar{j}+2 \bar{k})+t(4 \bar{i}+3 \bar{j}-\bar{k})$ and $\overline{r}=(\bar{i}+2 \bar{j}-4 \bar{k})+s(6 \bar{i}+3 \bar{j}-2 \bar{k})$ is

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ is equal to -

The foot of the perpendicular from the point $(1, 2, 3)$ on the line $\vec{r} = (6 \hat{i} + 7 \hat{j} + 7 \hat{k}) + \lambda(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ has the coordinates: