$A(2,3,4), B(4,5,7), C(2,-6,3), D(4,-4, k)$ are four points. If the line $\overline{AB}$ is parallel to $\overline{CD}$,then $k$ is equal to

Find the angle between the two lines $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and $\frac{x-1}{10}=\frac{y+3}{2}=\frac{z+4}{-11}$.

The distance of the point $(2, 4, 0)$ from the point of intersection of the lines $\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1}$ and $\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2}$ 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 find the value of $k$.

The line $L$ passes through the point $(1, 2, 3)$. The distance of any point on the line $L$ from the line $\vec{r} = (-1, 3, 4) + \lambda(3, -2, 1)$ is constant. Then the line $L$ does not pass through the point:

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