If $\vec{r}=\hat{i}+\hat{j}+t(2 \hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+s(3 \hat{i}-5 \hat{j}+2 \hat{k})$ are the vector equations of two lines $L_1$ and $L_2$, then the shortest distance between them is

If the harmonic conjugate of $P(2, 3, 4)$ with respect to the line segment joining the points $A(3, -2, 2)$ and $B(6, -17, -4)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma =$

The Cartesian equation of the line passing through the point $(-3,0,1)$ and perpendicular to vectors $\hat{i}-2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ 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 angle between two lines $\frac{x + 1}{2} = \frac{y + 3}{2} = \frac{z - 4}{-1}$ and $\frac{x - 4}{1} = \frac{y + 4}{2} = \frac{z + 1}{2}$ is

If the length of the perpendicular from the point $P(\beta, 0, \beta) \, (\beta \neq 0)$ to the line $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{-1}$ is $\sqrt{\frac{3}{2}}$,then $\beta$ is equal to