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

$A$ plane meets the coordinate axes at $P, Q,$ and $R$ such that the position vector of the centroid of $\Delta PQR$ is $2i - 5j + 8k$. Then the equation of the plane is:

Let $\pi_1$ be the plane passing through the points $(0,1,2), (1,0,-2), (-2,1,0)$ and $\pi_2$ be the plane passing through the point $(1,2,3)$ and perpendicular to the planes $x+y+z=1$ and $2x-3y+z=5$. If $\theta$ is the acute angle between the planes $\pi_1$ and $\pi_2$,then $\cos \theta=$

The Cartesian equation of a plane parallel to the plane $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$ and at a distance of $2$ units from it is

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the $X$-axis at $A$,$Y$-axis at $B$,and $Z$-axis at $C$. The area of $\triangle ABC$ is:

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angle between them:
$4x + 8y + z - 8 = 0$ and $y + z - 4 = 0$