If the line joining the points $(k, 3, 4)$ and $(4, 7, 8)$ is parallel to the line joining the points $(-1, -2, 1)$ and $(1, 2, l)$,then $k + l =$

Let the line of the shortest distance between the lines $L_1: \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $L_2: \vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})$ intersect $L_1$ and $L_2$ at $P$ and $Q$ respectively. If $(\alpha, \beta, \gamma)$ is the midpoint of the line segment $PQ$,then $2(\alpha+\beta+\gamma)$ is equal to . . . . . . .

The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{6-z}{2}$. Find the vector equation of the line.

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:

The shortest distance between the lines $\bar{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$ and $\bar{r}=(p+1) \hat{i}+(2 p-1) \hat{j}+(2 p+1) \hat{k}$ is

The length of the perpendicular drawn from the origin to the line $\bar{r} = (4\hat{i} + 2\hat{j} + 4\hat{k}) + \lambda(3\hat{i} + 4\hat{j} - 5\hat{k})$ is .......