The foot of the perpendicular drawn from the origin to the plane is $(4, -2, 5)$. Then,the Cartesian equation of the plane is:

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

The angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$ is

The position vectors of two points $P$ and $Q$ are $3i + j + 2k$ and $i - 2j - 4k$ respectively. The equation of the plane passing through $Q$ and perpendicular to $PQ$ 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 locus represented by $xy + yz = 0$ is