Explain parallel planes and project their separation

Their normals satisfy $(4,2,-4)=2(2,1,-2)$, so the normal directions are parallel. Divide the entire second equation by 2: it becomes $\mathbf r\cdot(2,1,-2)=-9/2$, with a different constant from 3. Thus the planes are distinct.

Following the Teacher projection method, choose $A=(0,3,0)$ on $p_1$ and $B=(0,-9/2,0)$ on $p_2$. Then

$\overrightarrow{BA}=A-B=(0,\tfrac{15}{2},0).$

Use the unit normal $\widehat{\mathbf n}=(2,1,-2)/3$:

$\begin{aligned}h&=\left|\overrightarrow{BA}\cdot\widehat{\mathbf n}\right|\\&=\frac13\left|(0,\tfrac{15}{2},0)\cdot(2,1,-2)\right|\\&=\frac52\text{ units}.\end{aligned}$

BA need not itself be perpendicular to the planes; its normal projection gives their separation. Equivalently, after matching the normals the distance is $|3-(-9/2)|/3=5/2$. Subtracting the original constants 3 and −9 before normalizing would be incorrect.