The shortest distance between the lines, where the first line passes through $(0,0,0)$ and $(2,0,3)$ and the second line passes through $(2,5,0)$ and $(0,4,0)$ is

The co-ordinates of the foot of the perpendicular from the origin to the plane $2x - 3y - 6z = 49$ are...

Let a plane P pass through the point $(3, 7, -7)$ and contain the line $\dfrac{x - 2}{-3} = \dfrac{y - 3}{2} = \dfrac{z + 2}{1}$. If the distance of the plane P from the origin is d, then $d^2$ is

The plane $\dfrac{x}{2} + \dfrac{y}{3} + \dfrac{z}{4} = 1$ cuts the co-ordinate axes at the points A, B, C respectively. Then the area of triangle ABC is