I'm looking for a proof of the following theorem:
If P1 written as $A = px + qy + rz +d = 0$ and P2 written as $B = p'x + q'y + r'z + d' = 0$
All the planes going through the line intersection of P1 and P2 can be written as: $$A+mB = 0 \ \text{ or }\ B + mA = 0.$$ It means that the plane A + mB = 0 and P1 and P2 have the same line intersection.
Example: Find the plane going through $(1,4,2)$ And the intersection of
$P1: 2x+3y-z=6$
$P2: 3x-2y+z=0$
$2x+3y-z-6+m(3x-2y+z)=0 $ placing the point in this equation we get $m=2$ So the plane is $2x+3y-z-6+2(3x-2y+z)=0 $ Or $8x-y+z=6$
As f5r5e5d says, you can't get plane $P_2$ this way. Clearly each equation $A+mB=0$ is satisfied by all points on $L=P_1\cap P_2$. Let $P_3$ be a plane, different from $P_1$ and $P_2$ through line $L$. Let $Q$ be a point on $P_3$ but not on $L$. Then $Q$ is neither on $P_1$ nor on $P_2$. So $A(Q)\ne0$ and $B(Q)\ne0$. (Here $A(Q)$ is what you get when you put in the coordinates of $Q$ into $A=px+qy+rz+d$ etc.) Take $m=-A(Q)/B(Q)$. Then $Q$ lies on $A+mB=0$ so that's an equation for $P_3$.