The shortest distance between the lines $\overline{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})$ and $\overline{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} - 5\hat{k})$ is:

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 straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is

The angle between the lines $\vec{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-3}{2}$ is

Find the point of intersection of the lines $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1}$ and $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$.

Find the angle between the two lines $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and $\frac{x-1}{10}=\frac{y+3}{2}=\frac{z+4}{-11}$.