The angle between the lines $2x = 3y = -z$ and $6x = -y = -4z$ is (in $^{\circ}$)

The shortest distance between the lines $r = (3i - 2j - 2k) + t(i)$ and $r = (i - j + 2k) + s(j)$ ($t$ and $s$ being parameters) is

The coordinates of the point where the line joining the points $(3, 5, -7)$ and $(-2, 1, 8)$ is intersected by the $yz$-plane are given by:

$L_1$ is a line passing through the points with position vectors $\hat{i}-2 \hat{j}-\hat{k}$ and $4 \hat{i}-3 \hat{k}$. $L_2$ is a line passing through the points with position vectors $\hat{i}+2 \hat{j}-\hat{k}$ and $2 \hat{i}-4 \hat{j}-5 \hat{k}$. Then the distance between $L_1$ and $L_2$ is

If $\vec{r}=\hat{i}+\hat{j}+t(2 \hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+s(3 \hat{i}-5 \hat{j}+2 \hat{k})$ are the vector equations of two lines $L_1$ and $L_2$, then the shortest distance between them is

The vector equation of a line whose Cartesian equations are $y=2$ and $4x-3z+5=0$ is