Connect AB, and extend the intersection X axis to point P, which is the point on the X axis that maximizes the value of |PA-PB| according to the trilateral relationship of the triangle.

Point b is the midpoint of the square,

∴ Point P is the point on the AB extension line, and at this time, P (3 3,0) means op = 3；;

Let point A', the symmetry point of point A relative to Y axis, and connect point A'B with Y axis at point Q, then point A'B is the minimum value of QA+QB.

∫A′(- 1，2)，B(2， 1)，

Let the straight line through A'B be y=kx+b, then 2=-k+b 1=2k+b,

The solution is k=- 1 /3 b=5 /3,

∴ q (0 0,5/ 3), that is, OQ=5 /3,

∴OP？ OQ=3×5 /3 =5。

So the answer is: 5.