The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is

Let $Q$ be the cube with the set of vertices $\{(x_1, x_2, x_3) \in \mathbb{R}^3: x_1, x_2, x_3 \in \{0,1\}\}$. 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 $(0,0,0)$ and $(1,1,1)$ is in $S$. For lines $\ell_1$ and $\ell_2$,let $d(\ell_1, \ell_2)$ denote the shortest distance between them. Then the maximum value of $d(\ell_1, \ell_2)$,as $\ell_1$ varies over $F$ and $\ell_2$ varies over $S$,is

The shortest distance between the lines $\overline{r}=(3 \bar{i}-5 \bar{j}+2 \bar{k})+t(4 \bar{i}+3 \bar{j}-\bar{k})$ and $\overline{r}=(\bar{i}+2 \bar{j}-4 \bar{k})+s(6 \bar{i}+3 \bar{j}-2 \bar{k})$ is

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to ....................

The equation of a line passing through the point $(2, -1, 1)$ and parallel to the line whose equation is $\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{-3}$ is:

Let the values of $p$,for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{r}=(p\hat{i}+2\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$ is $\frac{1}{\sqrt{6}}$,be $a$ and $b$ $(a < b)$. Then the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is: