The shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ is

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 $\cos \theta=$

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

Let $D$ be the foot of the perpendicular drawn from the point $A(2,0,3)$ to the line joining the points $B(0,4,1)$ and $C(-2,0,4)$. Then the ratio in which $D$ divides $BC$ is

The Cartesian equation of a line is $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z-5}{5}$,then the vector equation of the line is

Prove that the lines $x=p y+q, z=r y+s$ and $x=p^{\prime} y+q^{\prime}, z=r^{\prime} y+s^{\prime}$ are perpendicular if $p p^{\prime}+r r^{\prime}+1=0$.