If the two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ have a point in common,then $k=$

The image of the point $ (1,6,3) $ in the line $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} $ is

The foot of the perpendicular from the point $(1, 2, 3)$ on the line $\vec{r} = (6 \hat{i} + 7 \hat{j} + 7 \hat{k}) + \lambda(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ has the coordinates:

$L_1$ is a line passing through the points with position vectors $\hat{i}-2 \hat{j}-\hat{k}$ and $4 \hat{i}-3 \hat{k}$. $L_2$ is a line passing through the points with position vectors $\hat{i}+2 \hat{j}-\hat{k}$ and $2 \hat{i}-4 \hat{j}-5 \hat{k}$. Then the distance between $L_1$ and $L_2$ 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 $k = . . . . .$.

The coordinates of the foot of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{6 - x}{-3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ are: