$3$d, geometry, planes. In a question, we are asked to

Find a plane in $\Bbb R^3$ that is perpendicular to each of the planes:
$2x+3y-2z=5$ and $x+2y-3z=8$.

My question is: if this third plane is supposed to be perpendicular to both of those planes, shouldn't both of those two planes be either mutually perpendicular or parallel?
If we find the dot product of the normal vectors of those planes, it isn't $0$, and the normal vectors aren't proportionate either (to be parallel).
Could someone help me understand where I've gone wrong?
 A: Comment:
Are not your door hinge planes perpendicular to the floor?
A: Two planes $z=0$ and $z=2x$ aren't mutually perpendicular or parallel, yet the plane $y=0$ is perpendicular to each of them. We don't require the three planes to be mutually perpendicular to each other, we just want one of them to be perpendicular with rest two.
For the question, one of the normal vector for both of planes is each $(2, 3, -2)$ and $(1, 2, -3)$. Say a normal vector of the plane you want to find is $(a, b, c)$. Then, we have both
$$(2, 3, -2)\cdot (a, b, c)=2a+3b-2c=0$$
$$(1, 2, -3)\cdot (a, b, c)=a+2b-3c$$
Solving the system, we get $a=5k, b=-4k, c=-k$. So, a normal vector for the plane we are looking is $(5, -4, -1)$. Any plane in the form  of $5x-4y-z=m$ where $m$ is any real number is a plane that satisfies the conditions.
A: the only condition is to impose the dot product between the normal vectors equal to $0$.
Let $v=(2,3,-2)^{T}$ the normal vector of the first plane; $w=(1,2,-3)^{T}$ the normal vector of the second plane and $t=(a,b,c)^{T}$ the normal vector of the unknown plane.
We impose:
\begin{equation}
\begin{cases}
v \cdot t=0 \\
w \cdot t=0
\end{cases}
\end{equation}
\begin{equation}
\begin{cases}
2a+3b-2c=0 \\
a+2b-3c=0
\end{cases}
\end{equation}
Then you solve the system and extract a base and you have the normal vector of the unknown plane.
