Find the length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$.

The Cartesian equation of the line which is parallel to the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ and passing through the point $(1, -3, 5)$ is:

Point $A$ lies at a distance of $6$ units from the point $(1, 0, 1)$ on the line $\frac{x - 1}{2} = \frac{y}{2} = \frac{z - 1}{1}$ in the negative $z$-direction. Find the coordinates of $A$.

The distance between the lines $\frac{x}{2}=\frac{y}{-1}=\frac{z}{2}$ and $\frac{x-1}{2}=\frac{y-1}{-1}=\frac{z-2}{2}$ is

The distance of the point $Q(0, 2, -2)$ from the line passing through the point $P(5, -4, 3)$ and perpendicular to the lines $\overrightarrow{r} = (-3 \hat{i} + 2 \hat{k}) + \lambda(2 \hat{i} + 3 \hat{j} + 5 \hat{k}), \lambda \in R$ and $\overrightarrow{r} = (\hat{i} - 2 \hat{j} + \hat{k}) + \mu(-\hat{i} + 3 \hat{j} + 2 \hat{k}), \mu \in R$ is

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is