Find the values of $p$ so that the lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

The distance between a point $P$ whose position vector is $5 \hat{i}+\hat{j}+3 \hat{k}$ and the line $r=(3 \hat{i}+7 \hat{j}+\hat{k})+t(\hat{j}+\hat{k})$ is

The equation of a line passing through the point $(2, -1, 1)$ and parallel to the line joining the points $\hat{i} + 2\hat{j} + 2\hat{k}$ and $-\hat{i} + 4\hat{j} + \hat{k}$ is

If the harmonic conjugate of $P(2, 3, 4)$ with respect to the line segment joining the points $A(3, -2, 2)$ and $B(6, -17, -4)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma =$

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 $\overline{r}=\overline{p}+s \overline{q}$.

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $x-3=\frac{y-k}{2}=z$ intersect,then the value of $k$ is