The distance between the parallel lines $\frac{x-2}{2}=\frac{y-4}{5}=\frac{z-1}{2}$ and $\frac{x-1}{3}=\frac{y+1}{5}=\frac{z+3}{2}$ is

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

The square of the distance of the point $P(5,6,7)$ from the line $\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}$ is equal to:

The lines $\frac{x - 1}{2} = \frac{y + 1}{2} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 6}{2} = \frac{z}{1}$ intersect each other at point

Let the point $A$ be the foot of the perpendicular drawn from the point $P(a, b, 0)$ to the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}$. If the midpoint of the line segment $PA$ is $(0, \frac{3}{4}, -\frac{1}{4})$, then the value of $a^2+b^2+\alpha^2$ is equal to:

The line passing through the points $(a, 1, 6)$ and $(3, 4, b)$ crosses the $yz$-plane at $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$. Then the value of $(3a + 4b)$ is: