The angle between the lines $3x = 2y = -z$ and $-x = 6y = -4z$ is

If $A(1,1,2)$, $B(4,2,1)$ and $C(2,3,5)$ are the vertices of a triangle, then a vector representing the median of the triangle through $A$ is

The equation of the line passing through the point $Q(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is

The distance of the point having position vector $-\hat{i} + 2\hat{j} + 6\hat{k}$ from the straight line passing through the point $(2, 3, -4)$ and parallel to the vector $6\hat{i} + 3\hat{j} - 4\hat{k}$ is

Let in a $\triangle ABC$,the length of the side $AC$ be $6$,the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ 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 $p p^{\prime}+r r^{\prime}+1=0$.