If the mirror image of the point $P(3, 4, 9)$ in the line $\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$,then $14(\alpha+\beta+\gamma)$ is :

The equation of the straight line passing through the points $(4, -5, -2)$ and $(-1, 5, 3)$ is

Let a line $L$ be perpendicular to both the lines $L_1: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$ and $L_2: \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}$. If $\theta$ is the acute angle between the lines $L$ and $L_3: \frac{x-7}{2} = \frac{y-7}{1} = \frac{z}{2}$, then $\tan \theta$ is equal to:

The angle between a line with direction ratios $2, 2, 1$ and the line joining the points $(3, 1, 4)$ and $(7, 2, 12)$ is

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 shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$ is (in $\sqrt{3}$)