The shortest distance between the lines $\frac{x+7}{-6}=\frac{y-6}{7}=\frac{z}{1}$ and $\frac{7-x}{2}=y-2=z-6$ is

If for some $\alpha \in R$,the lines $L_1: \frac{x+1}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $L_2: \frac{x+2}{\alpha}=\frac{y+1}{5-\alpha}=\frac{z+1}{1}$ are coplanar,then the line $L_2$ passes through the point

Find the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$.

If the square of the shortest distance between the lines $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}$ and $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}$ is $\frac{m}{n}$,where $m, n$ are coprime numbers,then $m+n$ is equal to:

$A$ line $L$ passes through points $A(1, 3, 2)$ and $B(2, 2, 1)$. If the mirror image of point $P(1, 1, -1)$ in the line $L$ is $(x, y, z)$,then $x+y+z=$

Find the equation of the line passing through the point $P(2, -3, 1)$ and perpendicular to the line $L: \frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1}$.