Find the angle between the two lines $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and $\frac{x-1}{10}=\frac{y+3}{2}=\frac{z+4}{-11}$.

The direction cosines of two lines are connected by the relations $l+m-n=0$ and $lm-2mn+nl=0$. If $\theta$ is the acute angle between those lines,then $\cos \theta=$

The point of intersection of the lines $\frac{x - 5}{3} = \frac{y - 7}{-1} = \frac{z + 2}{1}$ and $\frac{x + 3}{-36} = \frac{y - 3}{2} = \frac{z - 6}{4}$ is

The distance of the point $Q(0, 2, -2)$ from the line passing through the point $P(5, -4, 3)$ and perpendicular to the lines $\overrightarrow{r} = (-3 \hat{i} + 2 \hat{k}) + \lambda(2 \hat{i} + 3 \hat{j} + 5 \hat{k}), \lambda \in R$ and $\overrightarrow{r} = (\hat{i} - 2 \hat{j} + \hat{k}) + \mu(-\hat{i} + 3 \hat{j} + 2 \hat{k}), \mu \in R$ is

Find the vector equation of the line $\frac{x - 2}{2} = \frac{2y - 5}{-3}, z = -1$.

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