No, it does not. Let’s suppose Alice’s pre-negotiated (or predetermined by a seeded PRNG) choices for rows 1..3 are:

(A11 A12 A13)

(A21 A22 A23)

(A31 A32 A33)

And Bob’s choices for columns 1..3 are:

(B11) (B21) (B13)

(B21) (B22) (B23)

(B31) (B32) (B33)

If they are given row 1 and column 1, the intersect should match, i.e. A11 == B11. The same goes for any row and column selection: Aij == Bij. That is not, however, consistent with the requirement that the sum of all numbers for Alice must be an odd number (each row must add up to an odd number and there are 3 rows), and the sum of all numbers for Bob must be an even number (each column must add up to an even number and the sum of three even numbers is even). So, that’s the real catch for the classical setup. And yes, quantum correlation can beat it.