The shortest distance between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ lies in the interval

$A$ 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 $x+y+z=$

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

$A$ line $L_1$ passes through the point with position vector $3 \hat{i}$ and is parallel to the vector $-\hat{i}+\hat{j}+\hat{k}$. Another line $L_2$ passes through the point with position vector $\hat{i}+\hat{j}$ and is parallel to the vector $\hat{i}+\hat{k}$. Find the position vector of the point of intersection of lines $L_1$ and $L_2$.

The distance of the point $P(7, 10, 11)$ from the line $\frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3}$ along the line $\frac{x-9}{2} = \frac{y-13}{3} = \frac{z-17}{6}$ is

If points $P(4, 5, x)$,$Q(3, y, 4)$,and $R(5, 8, 0)$ are collinear,then the value of $x+y$ is