Two lines $L_1: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_2: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value$(s)$.

If the lines $\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$ intersect,find the value of $k$.

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:

The lines $\overrightarrow{r} = (\hat{i} - \hat{j}) + \ell(2\hat{i} + \hat{k})$ and $\overrightarrow{r} = (2\hat{i} - \hat{j}) + m(\hat{i} + \hat{j} - \hat{k})$:

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

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,what is the angle between them?