The angle between the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ is

$ABC$ is a triangle in a plane with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $\lambda + \mu$ is:

Consider the lines $L_1$ and $L_2$ given by
$L_1: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2}$
$L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}$
$A$ line $L_3$ having direction ratios $1, -1, -2$ intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively. Then the length of line segment $PQ$ is

The shortest distance between the lines $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$ and $x+y+z+1=0, 2x-y+z+3=0$ is

If the line passing through the points $(-5, 1, 3)$ and $(1, 2, 0)$ is perpendicular to the line passing through the points $(x, 2, 1)$ and $(0, -4, 6)$,then $x = \dots$

$A$ 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 $L_2$.