$A$ plane $\pi$ is passing through $(2,0,1)$ and $(3,-3,4)$ and is perpendicular to the plane $x-2y+z=6$. Among the following,the plane that is perpendicular to the plane $\pi$ is:

$A$ plane passes through the point $(3, 5, 7)$. If the direction ratios of its normal are equal to the intercepts made by the plane $x+3y+2z=9$ with the coordinate axes,then the equation of that plane is

If a plane passes through $(1, -2, 1)$ and is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,then the distance of that plane from the point $(1, 2, 2)$ is

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:

Let three planes be:
$P_1 : x - y + z = 1$
$P_2 : x + y - z = -1$
$P_3 : x - 3y + 3z = 2$
Let $L_1, L_2, L_3$ be the lines of intersection of planes $(P_2, P_3)$,$(P_3, P_1)$,and $(P_1, P_2)$ respectively.
Statement-$1$: At least two of the lines $L_1, L_2, L_3$ are not parallel.
Statement-$2$: The three planes do not have a common point.

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x + y - 2z = 18$ are