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Find the constraint on the area of the field. Let the area of the field be \( \mathrm{S} \) hectares, then:
\[ 10 \cdot 300 \cdot S \leq 10000 \]
\[ \mathrm{S} \leq \frac{10}{3} \]

Note that the area of the field cannot be zero, consists of one part, and not multiple sections. The field is symmetric with respect to the coordinate axes for any parameter value.

Let \( 2a = -b \), then:
\[ a = -\frac{b}{a} \]

Consider \( |6|y| + 3| \); the value of this expression is obviously not less than three.

Thus, \( 2|2|x|-b| \) is at most 9, and therefore:
\[ \mathrm{b} \geq -4.5 \]

For \( b > 4.5 \), the point \( (0, 0) \) is not inside the bounded area, which must be a single region. Therefore, the constraint is from \(-4.5\) to \(4.5\). As we need integer values, we have the following parameter set: \( \pm 4, \pm 3, \pm 2, \pm 1, 0 \).

Testing in ascending order:
1. Let \( \mathrm{b} = -4 \)
\[ 2(2x + 4) + (6y + 3) = 12 \]
\[ \mathrm{x} = 0, \mathrm{y} = \frac{1}{6} \]
\[ \mathrm{y} = 0, \mathrm{x} = \frac{1}{4} \]
\[ s = 2 \cdot \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{12} < \frac{10}{3}, a = 2 \]

2. Let \(\mathrm{b} = -3\)
\[ 4x + 6 + 6y + 3 = 12 \]
\[ 4x + 6y = 3 \]
\[ \mathrm{x} = 0, \mathrm{y} = \frac{1}{2} \]
\[ \mathrm{y} = 0, \mathrm{x} = \frac{3}{4} \]
\[ \mathrm{s} = 2 \cdot \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{4} < \frac{10}{3}, a = 1.5 \text{ (not an integer)} \]

3. Let \(\mathrm{b} = -2\)
\[ 4x + 4 + 6y + 3 = 12 \]
\[ 4x + 6y = 5 \]
\[ \mathrm{x} = 0, \mathrm{y} = \frac{5}{6} \]
\[ \mathrm{y} = 0, \mathrm{x} = \frac{5}{4} \]
\[ s = 2 \cdot \frac{5}{6} \cdot \frac{5}{4} = \frac{25}{12} < \frac{10}{3}, a = 1 \]

4. Let \(\mathrm{b} = -1\)
\[ 4x + 2 + 6y + 3 = 12 \]
\[ 4x + 6y = 7 \]
\[ \mathrm{x} = 0, \mathrm{y} = \frac{7}{6} \]
\[ \mathrm{y} = 0, \mathrm{x} = \frac{7}{4} \]
\[ s = 2 \cdot \frac{7}{6} \cdot \frac{7}{4} = \frac{49}{12} > \frac{10}{3} \text{ (does not fit)} \]

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