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

Let $A$ be the point of intersection of the lines $L_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and $L_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let $B$ and $C$ be points on the lines $L_1$ and $L_2$ respectively such that $AB = AC = \sqrt{15}$. Then the square of the area of the triangle $ABC$ 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 $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{-1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then $k$ is equal to

The angle between a line with direction ratios $2: 2: 1$ and a line joining the points $(3, 1, 4)$ and $(7, 2, 12)$ is

The vector equation of the line $2x+4=3y+1=6z-3$ is