If two planes intersect,then the shortest distance between the planes is

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin $(0, 0, 0)$ to a plane,then the equation of that plane is:

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them: $2x - 2y + 4z + 5 = 0$ and $3x - 3y + 6z - 1 = 0$.

The perpendicular distance of the point $(1, -1, 2)$ from the plane $x + 2y + z = 4$ is

Let $\alpha x+\beta y+\gamma z=1$ be the equation of a plane passing through the point $(3, -2, 5)$ and perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 5)$. Then the value of $\alpha \beta \gamma$ is equal to $..........$.

The distance between two parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is given by $\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$. If the plane $2x-y+2z+3=0$ is at distances of $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4x-2y+4z+\lambda=0$ and $2x-y+2z+\mu=0$ respectively,then the maximum value of $\lambda+\mu$ is: