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

The equation of the plane passing through the points $(-1, -2, 0)$ and $(2, 3, 5)$ and parallel to the line $r = -3j + k + \lambda(2i + 5j - k)$ 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

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 coordinates of the foot of the perpendicular drawn from the origin to a plane is $(2, 4, -3)$. The equation of the plane is

$A$ plane $\pi$ passes through the points $(5,1,2)$,$(3,-4,6)$,and $(7,0,-1)$. If $p$ is the perpendicular distance from the origin to the plane $\pi$ and $l, m, n$ are the direction cosines of a normal to the plane $\pi$,then $|3l+2m+5n|=$