Find the intercepts cut off by the plane $2x + y - z = 5$.

If $A$ and $B$ are the feet of the perpendiculars drawn from the point $Q(a, b, c)$ to the planes $YZ$ and $ZX$ respectively,then the equation of the plane passing through the points $A, B$,and $O$ is (where $O$ is the origin).

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 passes through the points $A (1, 2, 3)$,$B (2, 3, 1)$,and $C (2, 4, 2)$. If $O$ is the origin and $P$ is $(2, -1, 1)$,then the projection of $\overline{OP}$ on this plane is of length .... .

If $P$ is the point $(2, 6, 3)$,then the equation of the plane passing through $P$ and perpendicular to $OP$,where $O$ is the origin,is:

Let $P$ be an arbitrary point such that the sum of the squares of the distances from the planes $x + y + z = 0$,$lx - nz = 0$,and $x - 2y + z = 0$ is equal to $9$. If the locus of the point $P$ is $x^2 + y^2 + z^2 = 9$,then the value of $l - n$ is equal to ......