The shortest distance between the lines $\dfrac{x - 3}{3} = \dfrac{y - 8}{-1} = \dfrac{z - 3}{1}$ and $\dfrac{x + 3}{-3} = \dfrac{y + 7}{2} = \dfrac{z - 6}{4}$ is

A plane meets the co-ordinate axes in A, B, C such that the centroid of the triangle ABC is the point $(1, r, r^2)$, then the equation of the plane is,

If M denotes the midpoint of the line joining A$(4, 5, -10)$ and B$(-1, 2, 1)$, then the equation of the plane through M and perpendicular to AB is:

The difference between the maximum value and the minimum value of the objective function $z = 3x + y$ subject to the constraints $2x + 3y \leq 6$, $x + y \geq 1$, $x \geq 0$, $y \geq 0$ is....