Hm, you seem to have some misconceptions. Let me try to get rid of some:
First of all, you seem to think the scientist would calculate the psi function directly at the detection plate. This is however wrong. To be clear: it is what he eventually wants, but he cannot get it so cheaply. It's good to think about the classical analog: if a scientist wants to calculate where a particle will be, it will not have an equation that it has to solve which immediately gives the prediction: he has to put in an initial condition!
Let me state it more clearly: in the case of the two-slit experiment, you don't know (a priori) where the system will eventually be, but you do know how it starts out: let's suppose that you shoot the particles horizontally with a fixed momentum p. Hence we know the initial wave function (classical analogy: we know the initial position and momentum). The wave function that is directed to the right and has a fixed momentum is proportional to e^{ipx / \hbar}. This gives us \psi(x,y,z,0): the initial condition.
Now this initial wave will evolve according to the Schrödinger equation (classical analogy: a particle with a certain position and momentum at time zero will evolve according to Newton's second law). Much like in classical mechanics we have to solve Newton's law to find the time evolution, we now have to solve i \hbar \frac{\partial}{\partial t} \psi(x,y,z,t) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi(x,y,z,t) + V(x,y,z) \psi(x,y,z,t). This equation is all you need.
This equation is fundamental. You also know another equation, where the i \hbar \frac{\partial}{\partial t} \psi(x,y,z,t) is replaced by E \psi(x,y,z,t).
That equation is not fundamental but just happens to be convenient for some purposes at some times, but for now: forget it!
What is important is that we have our initial condition, and our Schrödinger equation, which determines the time evolution of that wave function. With some math you thus get the correct wave function at all times. You then look at x = L (assuming the screen is at x = L) at a later time, when the wave has reached the screen, and then and only then (after taking the square modulus) have you obtained the probability distribution across the plate.
This finishes the double slit experiment.
Now what about the E \psi(x,y,z,t) equation?
Well, first remember the e^{ipx / \hbar} wave function. I said it represents a wave function with momentum p. Why did I say that? Well, we say that a wave has a certain value x of the observable X if the wave function is an eigenvector of the corresponding operator \hat X with eigenvalue x (I know that's a tricky sentence, but read it carefully: x, X, and \hat X are all different things, but intimately connected!). An example will help: let's denote the momentum observable by P (by "momentum observable", I mean the physical concepts of momentum). The
specific value I mentioned I called small p. Now the observable P has an associated operator that works on wave functions. For momentum it's \hat P = -i \hbar \frac{\partial}{\partial x} (keep it one-dimensional at the moment). This formula might look mysterious, but let's accept this formula at the moment, okay? It's a well-defined operator: if you give it a function, it returns another. Now again take my earlier sentence
a wave has a certain value x of the observable X if the wave function is an eigenvector of the corresponding operator \hat X with eigenvalue x
or applied to momentum:
a wave has a certain value momentum p if the wave function is an eigenvector of the corresponding operator \hat P= -i \hbar \frac{\partial}{\partial x} with eigenvalue p.
I.e. given a function psi we say it has a certain momentum p if \hat P \psi = p \psi (because this is by definition what it means to be an eigenvector of \hat P with eigenvalue p).
You see that e^{ipx/ \hbar} indeed satisfies this equation! Hence we say that this wave has momentum p.
Now back to the E \psi(x,y,z,t) equation! It's completely analogous! Above was for momentum, but we can do the same for energy:
We say that a wave function has energy E if it is an eigenvector of the operator \hat E with eigenvalue E
and traditionally \hat E is denoted by \hat H. The form of this operator (again: accept it for now!) \hat H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V. We say that a wave function has energy E if and only if E \psi = \hat H \psi. Analogous to the above with momentum!
Note that the Schrödinger equation can be rewritten as i \hbar \frac{\partial}{\partial t} \psi = \hat H \psi. So the fact that the Schrödinger equation looks so much like the E \psi equation is simply because it so happens that the operator corresponding to energy appears in the Schrödinger equation. But you see they're two very different equations: the Schrödinger equation is fundamental and determines the evolution of your system. The E \psi equation is simply used to find wave functions with certain energies, much like \hat P \psi = p \psi is used to find psi's with certain momenta (like e^{ipx/ \hbar}).
(It would perhaps be more logical to denote a value of energy by a small e instead of a big E, to be consistent, but again, tradition dictates we use a big letter for a value of energy, and luckily so, because it would be confusing with euler's constant!)
