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 $k = . . . . .$.

If a point $R(4, y, z)$ lies on the line joining the points $P(2, -3, 4)$ and $Q(8, 0, 10)$,then the distance of $R$ from the origin is

If the shortest distance between the lines $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$,then the sum of all possible values of $\lambda$ is :

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 $k =$

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

Let the line $L_{1}$ be parallel to the vector $-3\hat{i}+2\hat{j}+4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_{2}$ be parallel to the vector $2\hat{i}+\hat{j}+3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_{3}$ is parallel to the vector $-3\hat{i}+5\hat{j}+16\hat{k}$ and intersects the lines $L_{1}$ and $L_{2}$ at the points $C$ and $D$, respectively, then $|\overrightarrow{CD}|^2$ is equal to: