Find the equation of the plane that contains the point $(1, -1, 2)$ and is perpendicular to each of the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$.

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are three non-coplanar vectors, then the vector equation $\overrightarrow{r}=(1-p-q) \overrightarrow{a}+p \overrightarrow{b}+q \overrightarrow{c}$ represents a :

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$. Then,the equation of the plane is

$A$ point lying on the plane that passes through the points $\hat{i}-\hat{j}+\hat{k}$, $\hat{i}-2\hat{j}+3\hat{k}$, and $\hat{i}+2\hat{j}-3\hat{k}$ is:

If the foot of the perpendicular from $(0,0,0)$ to a plane is $(1,2,3)$, then the equation of the plane is

The direction cosines of the normal drawn to the plane passing through the points $(2,-1,5)$,$(1,-3,4)$,and $(5,2,1)$ are