The lines $\frac{6x-6}{18} = \frac{y+1}{3} = \frac{z-1}{5}$ and $\frac{3x+6}{12} = \frac{y-1}{3} = \frac{z+1}{2}$ are $\dots$

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

The foot of the perpendicular from the point $(1, 2, 3)$ on the line $\vec{r} = (6 \hat{i} + 7 \hat{j} + 7 \hat{k}) + \lambda(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ has the coordinates:

Let the vertices $Q$ and $R$ of the triangle $PQR$ lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$. Given $QR=5$ and the coordinates of the point $P$ are $(0,2,3)$. If the area of the triangle $PQR$ is $\frac{m}{n}$,then:

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.

The distance of the point $A(7, -2, 11)$ from the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$ measured along the line $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6}$ is: