The shortest distance between the lines $L_1: \bar{r} = \hat{i} + \hat{j} + \lambda(\hat{i} + \hat{j} - \hat{k})$ and $L_2: \bar{r} = \hat{j} + \hat{k} + \mu(\hat{j} + 2\hat{k} - \hat{i})$ is equal to:

Find the Cartesian equation of the line passing through the point $\hat{i} + 2\hat{j} + 2\hat{k}$ and parallel to the line joining the points $2\hat{i} - \hat{j} + \hat{k}$ and $-\hat{i} + 4\hat{j} + \hat{k}$.

The equations of motion of a rocket are: $x = 2t, y = -4t, z = 4t$ where the time $t$ is given in seconds,and the coordinates of a moving point in kilometers. What is the path of the rocket? At what distance will the rocket be from the starting point $O(0, 0, 0)$ in $10$ seconds?

If the shortest distance between the lines $\vec{r}_{1}=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in R, \alpha>0$ and $\vec{r}_{2}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k}), \mu \in R$ is $9$,then $\alpha$ is equal to $.....$

The equation of the line passing through the point $(2, 3, -4)$ and perpendicular to the $XOZ$ plane is

If $L_1$ is a line through the point $5 \hat{i}+8 \hat{j}+11 \hat{k}$ and parallel to the vector $2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $L_2$ is a line through the point $4 \hat{i}+6 \hat{j}+8 \hat{k}$ and parallel to the vector $3 \hat{i}+4 \hat{j}+5 \hat{k}$, then the point of intersection of $L_1$ and $L_2$ is