$P_{1c} = AP$ , $P_{2c} = BP$. How to find $P$? (being that $A$ and $B$ are $3\times 4$ matrices and $P$ is a $4\times 1$ vector) This problem arose in my stereo vision project.
$$
P_{1c} = A*P
$$
$$
P_{2c} = B*P
$$
where: $P_{1c}$ and $P_{2c}$ are $3\times1$ vectors, $A$ and $B$ are $3 \times 4$ matrices and $P$ is a $4\times1$ vector.  
Or, more verbosely:
$$
\left( 
\begin{array}{ccc}
a_1 \\
b_1 \\
w_1 \end{array} 
\right)
=
\left( 
\begin{array}{ccc}
    A_{11} & A_{12} & A_{13} & A_{14} \\
    A_{21} & A_{22} & A_{23} & A_{24} \\
    A_{31} & A_{32} & A_{33} & A_{34} \\
\end{array} 
\right)
*
\left(
\begin{array}{ccc}
    X \\
    Y \\
    Z \\
    1 \\
\end{array}
\right)
$$
$$
\left( 
\begin{array}{ccc}
a_2 \\
b_2 \\
w_2 \end{array} 
\right)
=
\left( 
\begin{array}{ccc}
    B_{11} & B_{12} & B_{13} & B_{14} \\
    B_{21} & B_{22} & B_{23} & B_{24} \\
    B_{31} & B_{32} & B_{33} & B_{34} \\
\end{array} 
\right)
*
\left(
\begin{array}{ccc}
    X \\
    Y \\
    Z \\
    1 \\
\end{array}
\right)
$$
$P_{1c}, P_{2c}, A$ and $B$ are known, find $P$.
An explicit solution is a nice start, and in fact I already have it:
(Explicit Solution - page 6) 
but from the nature of the problem, I can tell that it will not always be possible to find $P$ using the explicit solution (since I only have estimates for $P_{1c},P_{2c},A$ and $B$), so I was trying to tackle the problem like this:
$f(P) = (A*P-P_{1c})^2+(B*P-P_{2c})^2$
Minimize $f(P)$ with respect to $P$.
I believe this notation is not as rigorous as one would expect, since $P$ is a vector. So, to put it in another way:
Knowing $P_{1c},P_{2c},A$ and $B$, find the vector $P$ wich gives the "best-fit" to 
$$
P_{1c} = A*P
$$
$$
P_{2c} = B*P
$$
 (in a least squares sense).
Any help would be great, and please feel free to approach this problem with a different strategy than the one I'm trying, but I would appreciate if someone could answer this without "opening up" the matrices (instead of using the $A_{ij}$'s and working out with the explicit equations, only use the matrices, and matrix operations like $A^T, A^{-1}$), if that's even possible.
Hope I made myself clear, thank's in advance!

You have one matrix equation:
$$\pmatrix{P_{1c} \\ P_{2c}} = \pmatrix{A \\ B}\cdot P \tag{1}$$
It is an overdetermined system (unless there are too many dependencies in the data).
Here is what you want for the least squares solution to it:
$$P=\left[\pmatrix{A^\top & B^\top}\pmatrix{A \\ B}\right]^{-1}\pmatrix{A^\top & B^\top}\pmatrix{P_{1c} \\ P_{2c}}$$
There are methods that are more efficient than this explicit formula here, that avoids the extra calculation of squaring then inverting. You may notice that I got this using what is called a left inverse of $\pmatrix{A \\ B}$ on the equation (1). There are possibly more than one left inverse in general, but this one is the specific left inverse that gives the least squares solution.
Let $M = \pmatrix{A \\ B}$. Using user7530's helpful comment, it may be factored as $M=QR$, which is a standard form available in any computer package, simply called the QR factorization. I will show how to use that form later, but first the derivation of the formula I gave you.
To solve for $P$ in the equation $$X= MP \tag{2}$$ where $M$ has more rows than columns, consider a left inverse $\hat{M}$ such that $\hat{M}M=I$. At first glance, assuming the existence of $\hat{M}$, this seems to be the solution since it gives
$$\hat{M}X = \hat{M}MP = IP = P$$ and $$P=\hat{M}X$$ gives the solution. But the left inverse may not be unique. So the question is which left inverse do we want. Consider equation (2). We are looking for some $P$ to solve it, but it may not be possible. So instead we look at the closest possible solution. Any $P$ when right multiplied with $M$ will necessarily give a result that is in the column span of $M$, since a right multiply can only mix the columns of $M$. This along with one-sided inverses turns out to be all the considerations necessary to find the unique solution.
Since the result of $\hat{M}X$ should have the same column-span as $M$, this gives a unique $\hat{M}$ and it is

$$\hat{M} = \left[M^\top M\right]^{-1}M^\top$$

To use the QR factorization, we have $M=QR$. Then 
\begin{align}
P&=\hat{M} X \\
  &= \left[M^\top M\right]^{-1}M^\top X\\
 & = \left[R^\top Q^\top Q R\right]^{-1} M^\top X\\
 & = \left[R^\top  R\right]^{-1} M^\top X\\
\end{align}
Where the last line there is easier to compute since $R$ is triangular with zero rows.
