The angle between the lines $x-3y-4=0, 4y-z+5=0$ and $x+3y-11=0, 2y-z+6=0$ is

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$(s)$.

The shortest distance between the skew lines $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+t(\hat{i}+3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+s(2 \hat{i}+3 \hat{j}+\hat{k})$ is

Find the length of the perpendicular drawn from the point $P(3, -1, 11)$ to the line $\frac{x}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$.

The shortest distance between the lines $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}$ is equal to $............$

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