The equation of the plane passing through the point $(2, 3, 4)$ and parallel to the plane $x + 2y + 4z = 5$ is:

Find the distance between the point $P(6,5,9)$ and the plane determined by points $A(3,-1,2)$,$B(5,2,4)$,and $C(-1,-1,6)$.

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:

The distance between the planes given by $r \cdot (i + 2j - 2k) + 5 = 0$ and $r \cdot (i + 2j - 2k) - 8 = 0$ is

The equation of the plane passing through the point $(1, 2, -3)$ and perpendicular to the planes $3x + y - 2z = 5$ and $2x - 5y - z = 7$ is:

In three-dimensional space,the equation $3y + 4z = 0$ represents: