The distance of the point $P(7, 10, 11)$ from the line $\frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3}$ along the line $\frac{x-9}{2} = \frac{y-13}{3} = \frac{z-17}{6}$ is

Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{-1}$ and $\frac{x + 5}{7} = \frac{y - 2}{-6} = \frac{z - 3}{4}$ intersect at the point $R$. The reflection of $R$ in the $xy$-plane has coordinates

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:

Find the shortest distance between the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - \hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} - \hat{k}) + \mu(\hat{i})$.

Find the equation of a line parallel to the $x$-axis and passing through the origin.

The distance of the point $(2, 4, 0)$ from the point of intersection of the lines $\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1}$ and $\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2}$ is