Interaction of two Lines in 3D

Let $Q$ be the cube with the set of vertices $\left\{\left(x_1, x_2, x_3\right)\in {\mathrm{ℝ}}^3 : x_1, x_2, x_3\in \left\{0, 1\right\}\right\}$. Let $F$ be the set of all twelve lines containing the diagonals of the six faces of the cube $Q$. Let $S$ be the set of all four lines containing the main diagonals of the cube $Q$; for instance, the line passing through the vertices $\left(0, 0, 0\right)$ and $\left(1, 1, 1\right)$ is in $S$. For lines ${\ell }_1$ and ${\ell }_2$, let $d\left({\ell }_1, {\ell }_2\right)$ denote the shortest distance between them. Then the maximum value of $d\left({\ell }_1, {\ell }_2\right)$, as ${\ell }_1$ varies over $F$ and ${\ell }_2$ varies over $S$, is

Let $S$ be the set of all values of $\lambda$, for which the shortest distance between the lines $\frac{x-\lambda }{0}=\frac{y-3}{4}=\frac{z+6}{1}$and $\frac{x+\lambda }{3}=\frac{y}{-4}=\frac{z-6}{0}$ is $13$. Then $8\left|\sum _{\lambda \in S}\lambda \right|$ is equal to 

The shortest distance between the lines $\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$ is 

One vertex of a rectangular parallelopiped is at the origin $O$ and the lengths of its edges along $x\mathit{,}\mathit{ }y$ and $z$ axes are $3, 4$ and $5$ units respectively. Let $P$ be the vertex $(3, 4, 5).$ Then the shortest distance between the diagonal $OP$ and an edge parallel to $z$ axis, not passing through $O$ or $P$ is

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

Shortest distance between the lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ is

Shortest distance between the lines $\frac{x-5}{4}=\frac{y-3}{6}=\frac{z-2}{4}$ and $\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-9}{6}$ is

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.

Find the shortest distance between the two lines $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and $\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}$

The distance between two parallel lines $\overrightarrow{r}=\left(2\hat{i}+5\hat{j}+6\hat{k}\right)+\lambda \left(\hat{i}+\hat{j}\right)$ and $\overrightarrow{r}=\left(3\hat{i}+5\hat{j}+6\hat{k}\right)+\mu \left(\hat{i}+\hat{j}\right)$ is

The shortest distance between the lines $\frac{x+7}{-6}=\frac{y-6}{7}=z$ and $$$\frac{7-x}{2}=y-2=z-6$ is

The shortest distance between the line passing through the point $\hat{i}+2\hat{j}+3\hat{k}$ and parallel to the vector  $2\hat{i}+3\hat{j}+4\hat{k}$ and the line passing through the point $2\hat{i}+4\hat{j}+5\hat{k}$ and parallel to the vector $3\hat{i}+4\hat{j}+5\hat{k}$ is

Let two lines $L_1$ and $L_2$ be given by the vector equations $\overrightarrow{r}=\hat{i}+\hat{j}+\lambda (2\hat{i}-\hat{j}+\hat{k})$ and $\overrightarrow{r}=2\hat{i}+\hat{j}-\hat{k}+\mu (3\hat{i}-5\hat{j}+2\hat{k})$ respectively. The shortest distance between $L_1$ and $L_2$ is

The shortest distance between the lines given by

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

is equal to