Let the original length and breadth of the rectangular field be \( l \) and \( b \).
Given:
\[
l \times b = 260.
\]
When modified:
\[
(l - 5) = (b + 2).
\]
Thus,
\[
l - 5 = b + 2.
\]
\[
l - b = 7.
\]
Solving \( l \times b = 260 \) and \( l - b = 7 \):
Substituting \( l = b + 7 \) in the area equation:
\[
(b+7) \times b = 260.
\]
\[
b^2 + 7b - 260 = 0.
\]
Using the quadratic formula:
\[
b = \frac{-7 \pm \sqrt{49 + 1040}}{2} = \frac{-7 \pm 33}{2}.
\]
\[
b = 13, \quad l = 20.
\]
Thus, \( l = 20 \) meters, \( b = 13 \) meters.