The shortest distance between the skew lines $\vec{r}=(-\hat{i}-2 \hat{j}-3 \hat{k})+t(3 \hat{i}-2 \hat{j}-2 \hat{k})$ and $\vec{r}=(7 \hat{i}+4 \hat{k})+s(\hat{i}-2 \hat{j}+2 \hat{k})$ is

The equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to each of the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is:

The shortest distance between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ lies in the interval

If the position vectors of two points $P$ and $Q$ are respectively $9\hat{i} - \hat{j} + 5\hat{k}$ and $\hat{i} + 3\hat{j} + 5\hat{k}$,and the line segment $PQ$ intersects the $YOZ$ plane at a point $R$,then the ratio $PR : RQ$ is equal to

The straight lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 1}{2} = \frac{y - 2}{2} = \frac{z - 3}{-2}$ are

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