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

The points $A(3, 2, 0)$,$B(5, 3, 2)$,and $C(-9, 6, -3)$ are the vertices of a triangle $ABC$. If $AD$ is the internal bisector of $\angle BAC$ which meets $BC$ at $D$,then the coordinates of $D$ are:

If points $P(4, 5, x)$,$Q(3, y, 4)$,and $R(5, 8, 0)$ are collinear,then the value of $x+y$ 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 shortest distance between lines $\overline{r}=(2 \hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\overline{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+\hat{j}-5 \hat{k})$ is

Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$. The distance of the point $A\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ from the line $L$ along the line $\frac{3x-11}{2} = \frac{3y-11}{1} = \frac{3z-19}{2}$ is equal to: