The angle between the lines $\frac{x-3}{1}=\frac{y-2}{2}=\frac{z+4}{2}$ and $\frac{x-5}{3}=\frac{y+2}{2}=\frac{z}{6}$ 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 . . . . . . .

If the lines $\frac{1-x}{2}=\frac{7y+4}{2\lambda}=\frac{2z-5}{2}$ and $\frac{7-7x}{3\lambda}=\frac{y-1}{7}=\frac{6-z}{5}$ are at right angles,then the value of $\lambda$ is

The shortest distance between the skew lines $\vec{r}=(3 \hat{i}+4 \hat{j}-2 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})$ and $\vec{r}=(\hat{i}-7 \hat{j}-2 \hat{k})+\mu(\hat{i}+3 \hat{j}+2 \hat{k})$ is

The angle between the lines $\frac{x + 4}{1} = \frac{y - 3}{2} = \frac{z + 2}{3}$ and $\frac{x}{3} = \frac{y - 1}{-2} = \frac{z}{1}$ is

Find the vector equation of the line which is parallel to the vector $3 \hat{i}-2 \hat{j}+6 \hat{k}$ and which passes through the point $(1, -2, 3)$.