Find the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$.

Find the cartesian equation of the line which passes through the point $(-2, 4, -5)$ and is parallel to the line given by $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.

Line $L_1$ passes through the point $(1, 2, 3)$ and is parallel to the $z$-axis. Line $L_2$ passes through the point $(\lambda, 5, 6)$ and is parallel to the $y$-axis. Let for $\lambda = \lambda_1, \lambda_2$ with $\lambda_2 < \lambda_1$,the shortest distance between the two lines be $3$. Then the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $L_1$ is:

Points $P$ and $Q$ are given by $\vec{OP} = \hat{i} - \hat{j} - \hat{k}$ and $\vec{OQ} = -\hat{i} + \hat{j} + \hat{k}$. $A$ line along the vector $\vec{a} = \hat{i} + \hat{j}$ passes through the point $P$ and another line along the vector $\vec{b} = \hat{j} - \hat{k}$ passes through the point $Q$. If a line along the vector $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ intersects both the lines along the vectors $\vec{a}$ and $\vec{b}$ at $L$ and $M$ respectively,then $\vec{PM} =$

The angle between two lines $\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}$ and $\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}$ is

The shortest distance (in units) between the lines $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})$ is