If 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 perpendicular, then the value of $p$ is . . . . . . .

The shortest distance between lines $\bar{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\bar{r}=(2 \hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}-\hat{j}+\hat{k})$ is

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 point of intersection of the lines represented by $\overline{r}=(\overline{i}-6 \overline{j}+2 \overline{k})+t(\overline{i}+2 \overline{j}+\overline{k})$ and $\overline{r}=(4 \overline{j}+\overline{k})+s(2 \overline{i}+\overline{j}+2 \overline{k})$ is

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

Show that the points $(2,3,4), (-1,-2,1),$ and $(5,8,7)$ are collinear.