If the direction ratios of two lines are given by the equations $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the two lines is .....

The shortest distance between the lines $\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$ 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 line passing through the point $\bar{i} + 2\bar{j} + 3\bar{k}$ and parallel to the vector $2\bar{i} + 3\bar{j} + 4\bar{k}$ and the line passing through the point $2\bar{i} + 4\bar{j} + 5\bar{k}$ and parallel to the vector $3\bar{i} + 4\bar{j} + 5\bar{k}$ is:

If the shortest distance between the lines $\vec{r}_{1}=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in R, \alpha>0$ and $\vec{r}_{2}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k}), \mu \in R$ is $9$,then $\alpha$ is equal to $.....$

The shortest distance between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ lies in the interval