The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ 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 angle between the lines $\bar{r}=(3 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})$ and $\bar{r}=(5 \hat{i}-2 \hat{k})+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})$ is:

$A(-1, 2, -3), B(5, 0, -6), C(0, 4, -1)$ are the vertices of a triangle $ABC$. The direction cosines of the internal bisector of $\angle BAC$ are

The vector equation of the line whose Cartesian equations are $y=2$ and $4x-3z+5=0$ is

Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{-1}$ and $\frac{x + 5}{7} = \frac{y - 2}{-6} = \frac{z - 3}{4}$ intersect at the point $R$. The reflection of $R$ in the $xy$-plane has coordinates