If the vector equation of the plane $\bar{r}=(2 \hat{i}+\hat{k})+\lambda \hat{i}+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ in scalar product form is given by $\bar{r} \cdot(3 \hat{j}+2 \hat{k})=\alpha$,then $\alpha=$

The equation of the plane passing through $(1, 1, 1)$ and $(1, -1, -1)$ and perpendicular to $2x - y + z + 5 = 0$ is:

$A$ variable plane passes through the fixed point $(3, 2, 1)$ and meets $X, Y,$ and $Z$ axes at points $A, B,$ and $C$ respectively. $A$ plane is drawn parallel to the $YZ$-plane through $A$,a second plane is drawn parallel to the $ZX$-plane through $B$,and a third plane is drawn parallel to the $XY$-plane through $C$. Then the locus of the point of intersection of these three planes is:

Let the plane passing through the point $(-1, 0, -2)$ and perpendicular to each of the planes $2x + y - z = 2$ and $x - y - z = 3$ be $ax + by + cz + 8 = 0$. Then the value of $a + b + c$ is equal to:

Consider the following three planes:
$P : x + y - 2z + 7 = 0$
$Q : x + y + 2z + 2 = 0$
$R : 3x + 3y - 6z - 11 = 0$

The Cartesian equation of the plane $r = (1 + \lambda - \mu )i + (2 - \lambda )j + (3 - 2\lambda + 2\mu )k$ is