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

$A$ plane which is perpendicular to two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,passes through $(1, -2, 1)$. The distance of the plane from the point $(1, 2, 2)$ 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 Cartesian equation of the plane $\bar{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$ is

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

Find the equation of the plane passing through the points $(2, 1, 0)$,$(3, -2, -2)$,and $(3, 1, 7)$.