If the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda}$ and $\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$ are coplanar,then $\sin ^{-1}(\sin \lambda)+\cos ^{-1}(\cos \lambda)=$

The distance between the parallel lines $\frac{x-1}{2}=\frac{y-2}{-2}=\frac{z-3}{1}$ and $\frac{x}{2}=\frac{y}{-2}=\frac{z}{1}$ is

If the lines $\vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda(q\hat{i} - 2\hat{j} + \hat{k})$ and $\vec{r} = p\hat{i} - 3\hat{j} + 2\hat{k} + \mu(\hat{i} - 2\hat{j} + 2\hat{k})$ intersect each other and $q\hat{i} - 2\hat{j} + \hat{k}$ is collinear to $4\hat{i} - 4\hat{j} + 2\hat{k}$, then the values of $p$ and $q$ are

The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is

Let $L_1$ (respectively $L_2$) be the line passing through $2 \hat{i}-\hat{k}$ (respectively $2 \hat{i}+\hat{j}-3 \hat{k}$) and parallel to $3 \hat{i}-\hat{j}+2 \hat{k}$ (respectively $\hat{i}-2 \hat{j}+\hat{k}$). Then the shortest distance between the lines $L_1$ and $L_2$ is equal to

Find the length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$.