There're 2 parallel planes $P_1, P_2$ and the distance between them is 2. $P_1$ goes through points $A=(2,0,3)$ and $B=(0,0,6)$ and $P_2$ goes through point $C=(-2,0,2)$. Find the equations of the planes.
I thought of the following although the solution seems not correct and too long. First this is the visualization that I made and I hope is correct:
First we can find $\underline u$ the normal vector of vectors $BA$ and $BC$ through cross product which is:
$$\underline u=<2,0,-3> \times <-2,0,-4>=<0,14,0>$$
Now we can find $\underline n_1$ the normal vector of plane $P_1$ from: $$\underline n_1=\underline u \times BA=<0,14,0> \times <2,0,-3>=<42,0,28>$$ Now we have the plane equation of $P_1: 42x+28z+d=0$ after dividing by $14$ becomes: $3x+2z+{d \over 14}=0$.
I understand the normal vectors for $P_1$ and $P_2$ are the same but $d$'s are different. I could calculate the $d$ from the distance formula: $$ D=\frac{|3x+2z+d|}{\sqrt{3^2+2^2}}=2 $$
But then how would I find the $d$ for $P_2$?