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

Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

The shortest distance between the lines $\frac{x - 6}{1} = \frac{y - 2}{-2} = \frac{z - 2}{2}$ and $\frac{x + 4}{3} = \frac{y}{-2} = \frac{z + 1}{-2}$ is

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

Which of the following lines in ${R^3}$ is coincident with the line $\vec{r} = (1, 1, 1) + k(2, 3, 4), k \in R$?

If the shortest distance between the lines
$L_1: \overrightarrow{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, \lambda \in R$
$L_2: \overrightarrow{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, \mu \in R$
is $\frac{m}{\sqrt{n}}$,where $\operatorname{gcd}(m, n)=1$,then the value of $m+n$ equals.