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

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

Find the image of the point $(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.

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 distance between parallel lines $\vec{r}=(2\hat{i}-\hat{j}+\hat{k})+\lambda(2\hat{i}+\hat{j}-2\hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(2\hat{i}+\hat{j}-2\hat{k})$ is

$A$ line $L$ passes through points $A(1, 3, 2)$ and $B(2, 2, 1)$. If the mirror image of point $P(1, 1, -1)$ in the line $L$ is $(x, y, z)$,then $x+y+z=$