The distance of the plane $2x - 3y + 6z + 14 = 0$ from the origin is:

The equation of the plane,passing through the point $(-1, 2, -3)$ and parallel to the lines $\frac{x-1}{3} = \frac{y-2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y-1}{-3} = \frac{z-2}{2}$,is

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the axes at points $A, B,$ and $C$. Find the area of $\Delta ABC$.

The equation of the plane passing through the midpoint of the line segment joining the points $A(4, 5, -10)$ and $B(-1, 2, 1)$ and perpendicular to $AB$ is:

The plane which bisects the line segment joining the points $A(4, -2, 3)$ and $B(2, 4, -1)$ at right angles also passes through the point:

The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by