Lines $\frac{2x-5}{k} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other. Then,the value of $k$ is . . . . . . .

The vector equation of the line $2x+4=3y+1=6z-3$ is

The angle between the lines $\bar{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}+\hat{j}+2\hat{k})$ and $\bar{r}=(3\hat{i}+\hat{k})+\lambda^{\prime}(2\hat{i}+\hat{j}-\hat{k})$,where $\lambda, \lambda^{\prime} \in R$,is

The shortest distance between the lines $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}$ is equal to $............$

Let the foot of the perpendicular from the point $(\lambda, 2, 3)$ on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}$ is equal to:

The Cartesian equations of the line passing through the point $A(0, 1, 1)$ and parallel to the $X$-axis are: