The shortest distance between the lines $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ is (in $\sqrt{3}$)

Statement $1:$ The shortest distance between the lines $\frac{x}{2} = \frac{y}{-1} = \frac{z}{2}$ and $\frac{x-1}{4} = \frac{y-1}{-2} = \frac{z-1}{4}$ is $\sqrt{2}$.
Statement $2:$ The shortest distance between two parallel lines is the perpendicular distance from any point on one of the lines to the other line.

If the lines $\frac{x - 5}{5m + 2} = \frac{2 - y}{5} = \frac{1 - z}{-1}$ and $\frac{x}{1} = \frac{2y + 1}{4m} = \frac{1 - z}{-3}$ are perpendicular to each other, then the value of $m$ is ...

The length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$ is (in $\text{ units}$)

The Cartesian equation of the line passing through the point $(-3,0,1)$ and perpendicular to vectors $\hat{i}-2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ is

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: