The angle between the lines $2x = 3y = -z$ and $6x = -y = -4z$ is (in $^{\circ}$)

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

Let the values of $p$,for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{r}=(p\hat{i}+2\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$ is $\frac{1}{\sqrt{6}}$,be $a$ and $b$ $(a < b)$. Then the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is:

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

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then the value of $k$ is

Lines $\frac{x-5}{7}=\frac{y-5}{k}=\frac{z-2}{1}$ and $\frac{x}{1}=\frac{y-3}{2}=\frac{z+1}{3}$ are perpendicular to each other,then the value of $k=$ . . . . . . .