The measure of the angle between the lines $x = k + 1, y = 2k - 1, z = 2k + 3, k \in R$ and $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{1}$ is

If lines $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{6-z}{5}$ are perpendicular to each other,then $k=$ $\qquad$ .

If $A(3,-1,11)$,$B(0,2,3)$,and $C(4,8,11)$ are three points,then the coordinates of the foot of the perpendicular drawn from the point $A$ to the line joining the points $B$ and $C$ is

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 line passing through the point $\bar{i} + 2\bar{j} + 3\bar{k}$ and parallel to the vector $2\bar{i} + 3\bar{j} + 4\bar{k}$ and the line passing through the point $2\bar{i} + 4\bar{j} + 5\bar{k}$ and parallel to the vector $3\bar{i} + 4\bar{j} + 5\bar{k}$ is:

If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4},$ $l \neq 0$ is $(3,5,7),$ then the shortest distance between the line $L_{1}$ and line $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to: