Here is what I did. I considered 3 scenarios in which the third mass was just not in consideration. I established the motions of the masses in all three scenarios. Since I knew what the double derivative of that motion gave me, I substituted it in the equation of three bodies. I solved for them and got an answer. I did not equate the second part to zero. The second part just doesn't exist (for the time being). Then I substituted to get the answer

I get the motion of $M_1$ and $M_2$, right? I know that $x_1$ gives me the position vector of $M_1$. Also double derivative of $x_1$ gives me the acceleration vector. Since the same acceleration vector forms a component in the new acceleration, I can replace that acceleration with the double derivative of $x_1$.

Forget this. Suppose that I want to solve the differential equation. (The 3 body one). Clearly my solution does satisfy the differential equation and the initial conditions ( 6 initial conditions!). How do you explain that? ( all the $x_i$, $y_i$, $z_i$ are previously solved and I have just put them in my equation

@MathsGuy No, your solution doesn't satisfy the differential equation. You forget that $r(t)$ are different functions in the 3 vs 2 body problems when you make the substitutions. Try using the same "trick" on a similar system of 3 linear equations and you will immediately see where you went wrong.

@J. Delaney What is r(t) here? I didn't understand

Plus it satisfies. Double derivate $x_1$ to get G(m2)(R2-R1) ... Double derivate $z_1$ to get G(m2)(R3-R1). Both are vectors.

@J. Delaney I am assuming you are talking about function R(t)=R2 - R1. But as I have said in my earlier comment, by double derivating $x_1$ and $z_1$ you get the previous equation. Hence it satisfies. Plus all initial conditions are met

@MathsGuy what's the difference between $r_1$ and $x_1$ ?

If you are saying that $r_1$ and $x_1$ are same then I am ok with it.

Hello J. Delaney. we can talk from here from now on

Hello @J. Delaney. we can talk from here from now on

If they are the same then why did you use two different letters for them ?

Because I was going to use R(t) again later. I was having problem thinking of new but relevant letter

Is there something wrong with my method?

Yes. If you assume r1=x1 , you don't need additional symbols. Reformulate your analysis using only r1,r2,r3

Ok. Is there something wrong with my method of deriving the motions of the Three bodies?

If you'll write it only with r1,r2,r3 it will become immediately clear to you

First of all. Let us say that r1, r2, r3 are the position vectors of the 3 body situation. Let x1, x2, etc, be the position vectors of the 2 body situations. Putting x1,x2 etc does satisfy the differential equations. What are you talking about?

r1, r2, r3 are the position vectors of the 3 body situation