Shortest Distance between Two Lines

Find the distance of the point  $\left(-1, -5, -10\right)$, from the point of intersection of the line $\overrightarrow{r}=\left(2\widehat{i}-\widehat{j}+2\widehat{k}\right)+\lambda \left(3\widehat{i}+4\widehat{j}+2\widehat{k}\right)$ and the plane $\overrightarrow{r}\cdot \left(\widehat{i}-\widehat{j}+\widehat{k}\right)=5$.

Find the shortest distance between the lines $\overrightarrow{r}=3\hat{i}+2\hat{j}-4\hat{k}+\lambda (\hat{i}+2\hat{j}+2\hat{k})$and $\overrightarrow{r}=5\hat{i}-2\hat{j}+\mu (3\hat{i}+2\hat{j}+6\hat{k})$.

If the lines intersect find their point of intersection.

The shortest distances between the diagonals of a rectangular parallelepiped whose sides are $a, b, c$ and the edges not meeting it are $\frac{bc}{\sqrt{b^2+c^2}},\frac{ca}{\sqrt{c^2+a^2}},\frac{ab}{\sqrt{a^2+b^2}}$.

Shortest distance between the following pair of lines $\overrightarrow{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$ and $\overrightarrow{r}=(2\hat{i}-\hat{j}-\hat{k})+\mu (2\hat{i}+\hat{j}+2\hat{k})$ is

The following pairs of lines $\overrightarrow{r}=(\hat{i}-\hat{j})+\lambda (2\hat{i}+\hat{k})$ and $\overrightarrow{r}=(2\hat{i}-\hat{j})+\mu (\hat{i}+\hat{j}-\hat{k})$ are intersecting lines.

If the shortest distance between the lines $\overrightarrow{r}=(4\hat{i}-\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k})$ and $\overrightarrow{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu (\hat{i}+4\hat{j}-5\hat{k})$ is $\frac{1}{\sqrt{m}}$, then the value of $m$ is

If the shortest distance between the lines
$\overrightarrow{r}=(4\hat{i}-\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k})$and $\overrightarrow{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}+4\hat{j}-5\hat{k})$ is $\frac{a}{\sqrt{5}}$, then $a=$
 

Find the shortest distance between the pair of lines:

$\stackrel{\rightharpoonup }{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k}) and \stackrel{\rightharpoonup }{r}=3\hat{i}+3\hat{j}-5\hat{k}+\mu (2\hat{i}+3\hat{j}+6\hat{k})$

Find the shortest distance between the pair of lines:

$\frac{x-3}{2}=\frac{y-4}{1}=\frac{z+1}{-3}$ and $\frac{{\displaystyle x-1}}{{\displaystyle -1}}=\frac{{\displaystyle y-3}}{{\displaystyle 3}}=\frac{{\displaystyle z-1}}{{\displaystyle 2}}$

By computing the shortest distance determine whether the following lines intersect or not: $\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}$ and $\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}$.

By computing the shortest distance determine whether the following pairs of lines intersect or not: $\overrightarrow{r}=(\hat{i}-\hat{j})+\lambda (2\hat{i}+\hat{k})$ and $\overrightarrow{r}=(2\hat{i}-\hat{j})+\mu (\hat{i}+\hat{j}-\hat{k})$.

By computing the shortest distance, determine whether the lines $\overline{r}=(\hat{i}-\hat{j})+\lambda (2\hat{i}+\hat{k})$ and $\overline{r}=2\hat{i}-\hat{j}+\mu (\hat{i}-\hat{j}+\hat{k})$  intersect or not.

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then find the value of $k$.

If the shortest distance between the lines $\overrightarrow{r}=(\lambda -1)\hat{i}+(\lambda +1)\hat{j}-(1+\lambda )\hat{k}$ and $\overrightarrow{r}=(1-\mu )\hat{i}+(2\mu -1)\hat{j}+(\mu +2)\hat{k}$ is of the form of $\frac{a}{\sqrt{2}}$, then $a=$