The image of the point $ (1,6,3) $ in the line $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} $ is

If $A(3,4,5)$, $B(4,6,3)$, $C(-1,2,4)$, and $D(1,0,5)$ are points such that the angle between the lines $DC$ and $AB$ is $\theta$, then $\cos \theta$ is equal to

Assertion $(A)$: For the lines $\overline{r}=\overline{a}+t \overline{b}$ and $\overline{r}=\overline{p}+s \overline{q}$,if $(\bar{a}-\bar{p}) \cdot(\bar{b} \times \bar{q}) \neq 0$,then the two lines are coplanar.
Reason $(R)$: $|(\bar{a}-\bar{p}) \cdot(\bar{b} \times \bar{q})|$ is $|\bar{b} \times \bar{q}|$ times the shortest distance between the lines $\overline{r}=\overline{a}+t\bar{b}$ and $\overline{r}=\overline{p}+s \overline{q}$.

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

If the direction cosines of two lines are given by $l+m+n=0$ and $mn-2lm-2nl=0$,then the acute angle between those lines is

The coordinates of the point of intersection of the lines $\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 1}{-1}$ and $\frac{x - 4}{2} = \frac{y - 2}{-1} = \frac{z - 4}{2}$ are: