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Find the restriction on the field area. Let the area of the field be \( S \) hectares, then

\[ 
10 \times 300 \times S \leq 10000 
\]

\[ 
\mathrm{S} \leq \frac{10}{3} 
\]

Next, let’s note a few obvious points deriving from the problem statement: the field area cannot be zero, and the field consists of a single part, not several sections.

Also, note that the field is symmetric with respect to the coordinate axes for any parameter value, due to \( x \) and \( y \) being under the modulus.

Consider \( |6|y| + 3| \). The value of this expression is clearly not less than three.

Then, for \( 2|2|x| - \boldsymbol{a}| \), it must not exceed 9. Thus, obviously \( \boldsymbol{a} \geq -4.5 \). For smaller parameter values, the area does not exist.

Also, note that due to the absence of a parameter in the second modulus, the boundary stretches along the x-axis.

For \( a > 4.5 \), the point \( (0, 0) \) is not inside the bounded area, which is split in two parts, contrary to the logical sense of the problem.

Hence, the parameter ranges from \( -4.5 \) to \( 4.5 \). Considering that the area increases with increasing \( a \), as shown in illustrations, and that we need only integer values, we have the final set of parameters: \( \pm 4, \pm 3, \pm 2, \pm 1, 0 \).

Proceed in ascending order. Let \( a = -4 \).

\[ 
2(2x + 4) + (6y + 3) = 12 
\]

\[ 
\mathrm{x} = 0, \mathrm{y} = \frac{1}{6} 
\]

\[ 
\mathrm{y} = 0, \mathrm{x} = \frac{1}{4} 
\]

\[ 
\mathrm{s} = 2 \times \frac{1}{6} \times \frac{1}{4} = \frac{1}{12} < \frac{10}{3} 
\]

Let \( a = -3 \).

\[ 
4x + 6 + 6y + 3 = 12 
\]

\[ 
4x + 6y = 3 
\]

\[ 
\mathrm{x} = 0, \mathrm{y} = \frac{1}{2} 
\]

\[ 
\mathrm{y} = 0, \mathrm{x} = \frac{3}{4} 
\]

\[ 
\mathrm{s} = 2 \times \frac{1}{2} \times \frac{3}{4} = \frac{3}{4} < \frac{10}{3} 
\]

Let \( a = -2 \).

\[ 
4x + 4 + 6y + 3 = 12 
\]

\[ 
4x + 6y = 5 
\]

\[ 
\mathrm{x} = 0, \mathrm{y} = \frac{5}{6} 
\]

\[ 
\mathrm{y} = 0, \mathrm{x} = \frac{5}{4} 
\]

\[ 
\mathrm{s} = 2 \times \frac{5}{6} \times \frac{5}{4} = \frac{25}{12} < \frac{10}{3} 
\]

Let \( a = -1 \).

\[ 
4x + 2 + 6y + 3 = 12 
\]

\[ 
4x + 6y = 7 
\]

\[ 
\mathrm{x} = 0, \mathrm{y} = \frac{7}{6} 
\]

\[ 
\mathrm{y} = 0, \mathrm{x} = \frac{7}{4} 
\]

\[ 
\mathrm{s} = 2 \times \frac{7}{6} \times \frac{7}{4} = \frac{49}{12} > \frac{10}{3} 
\]

Already not suitable.

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