The distance between the two lines $L_1$ and $L_2$ given by $L_1: \dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{-z-4}{-6}$ and $L_2: \dfrac{x-3}{2} = \dfrac{y-3}{3} = \dfrac{z+5}{6}$ is

The points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ at a distance of $\sqrt{17}$ from the point (-2, -1, 3) are:

Find the root of perpendicular from origin to the line $\frac{x - 1}{3} = \frac{y - 2}{-2} = \frac{z + 1}{3}$.

The shortest distance between lines whose vector equations are: $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$, is:

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

$\vec{r} = 2\hat{i} - \hat{j} + \mu(\hat{i}-\hat{j}-\hat{k})$ is $\dfrac{\sqrt{14}}{14}m$ units, the value of $m$ is

Distance of the point $(2, 4, -1)$ from the line $\frac{10+2x}{2} = \frac{y+3}{4} = \frac{6-z}{9}$ is

The value of k for which the lines $\frac{2x - 3}{4} = \frac{3 - y}{k} = \frac{z - 2}{-2}$ and $\frac{x - 2}{1} = \frac{y}{4} = \frac{5 - z}{3}$ are perpendicular to each other is: