Privět, concerning the first problem, the minors don't change if the two 4-component rows of the $2 \times 4$ matrix are being subtracted from each other (like if you solve a set of linear equations) - because the 6 entries are really the components of a wedge products of the two 4-dimensional rows. In particular, without changing the minors, I can always bring the matrix to the form $\begin{array}{ccc} e&0&a&b\\0&1&c&d\end{array}$ The minors are then $e,ec,ed,-a,+b,ad-bc$. Note that I only have five parameters $a,b,c,d,e$ now to adjust six minors in your list. How do you figure out whether the six numbers may be written in this form? Note that in my list, once again, $e,ec,ed,-a,+b,ad-bc$ it's true that the last 6th entry multiplied the 1st one is equal to minus (the 4th times 3rd plus 5th times 2nd), $e(ad-bc)$. This formula looks unnatural but it must be possible to order the six minors as $A,B,C,D,E,F$ so that $AB+CD+EF=0.$ This is a nicely symmetric formula and there must exist a more covariant way to derive it, without putting zeros at some random places haha. Oh, I actually know how to derive the formula in a simple way. The 6 minors are the components of the wedge product $W = r_1\wedge r_2$ which belongs to a plane, so its wedge-square - a 4-form - has to vanish. Then it's enough to collect the coefficients in front of $e_1\wedge e_2\wedge e_3\wedge e_4$ of $W\wedge W$ - note that the latter doesn't vanish identically because they're two even 2-forms so the wedge product is symmetric. This coefficient will obviously look like $AB+CD+EF$.
At any rate, the equation above clearly can't be satisfied with positive minors. However, I am confident that with other positions of the zeros, and/or conventions for the signs of the minors, you may flip some signs in the equation above (two sign flips are equivalent to one - it's the only form of the equation inequivalent to the three identical terms), to get $AB+CD=EF$ In this form, it's straightforward to test the permutations of your three lists of minors. The possible values of $AB+CD-EF$, using all possible permutations of the six minors $A,B,C,D,E,F$, in the three cases are:
{-20, -19, -16, -11, -9, -5, -4, -1, 0, 4, 5, 7, 8, 11, 12, 13, 16, 17, 20, 23, 24, 25, 28, 29, 31, 32, 35, 36, 40, 43, 44, 45, 49, 52, 53, 56}
{-18, -17, -14, -7, -5, -1, 2, 5, 6, 10, 13, 14, 15, 19, 21, 22, 23, 25, 26, 30, 34, 35, 37, 38, 41, 43, 46, 47, 49, 50, 54, 58, 59, 61, 65, 70, 71, 74}
{-8, -7, -4, 7, 9, 13, 20, 23, 24, 28, 32, 35, 37, 41, 43, 47, 48, 49, 52, 56, 60, 61, 64, 65, 67, 71, 73, 77, 80, 83, 84, 88, 92, 97, 99, 103, 112, 113, 116}
Note that zero is only found in the first case, so only $(2,3,4,5,6,7)$ can be produced as minors of such a matrix. The Mathematica code I used:
a = {2, 3, 4, 5, 6, 7} + p {1, 1, 1, 1, 1, 1}; f[v_] := v[[1]]*v[[2]] + v[[3]]*v[[4]] - v[[5]]*v[[6]]; b = Permutations[a]; Union[Map[f, b] /. {p -> 0}] Union[Map[f, b] /. {p -> 1}] Union[Map[f, b] /. {p -> 3}]
If I extend the function $f$ above to
f[v_] := {v, v[[1]]*v[[2]] + v[[3]]*v[[4]] - v[[5]]*v[[6]]};
then the results will also show me the right permutation that produced zero: $ 2\times 5 + 3\times 6 - 4\times 7 = 0.$ Then it's straightforward to write down a matrix that has the right minors $\begin{array}{ccc} 2&0&1&-3/2\\0&2&3&5/2\end{array}$ The subdeterminants are $4,6,5,-2,3,7$ for the pairs of columns $12,13,14,23,24,34$.