The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ 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 $k=$ . . . . . . .

The cosine of the angle included between the lines $\overline{r}=(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}-2 \hat{k})$ and $\overline{r}=(\hat{\imath}+\hat{\jmath}+3 \hat{k})+\mu(3 \hat{\imath}+2 \hat{\jmath}-6 \hat{k})$ where $\lambda, \mu \in R$ is

The perpendicular distance from the point $P(3, -2, 1)$ to the line joining the points $A(1, -3, 5)$ and $B(2, 1, -4)$ is:

The point of intersection of the lines $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}$ is

The foot of the perpendicular from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$ is