To solve this optimization problem, we first need to convert the natural language description into a symbolic representation. Let's denote:

- $x_1$ as the number of individual servings,
- $x_2$ as the number of family servings.

The objective is to maximize profit. Given that each individual serving is sold for a profit of $3 and each family serving is sold for a profit of $10, the objective function can be represented algebraically as:

$$3x_1 + 10x_2$$

The constraints are as follows:
1. The total amount of soup used does not exceed 50,000 ml: Since an individual serving has 250 ml and a family serving has 1200 ml,
   $$250x_1 + 1200x_2 \leq 50000$$
2. The number of individual servings must be at least three times the number of family servings:
   $$x_1 \geq 3x_2$$
3. At least 10 family servings need to be made:
   $$x_2 \geq 10$$
4. Non-negativity constraints since the number of servings cannot be negative:
   $$x_1, x_2 \geq 0$$

Thus, the symbolic representation in JSON format is:

```json
{
    'sym_variables': [('x1', 'number of individual servings'), ('x2', 'number of family servings')],
    'objective_function': '3*x1 + 10*x2',
    'constraints': [
        '250*x1 + 1200*x2 <= 50000',
        'x1 >= 3*x2',
        'x2 >= 10',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

Now, let's write the Gurobi code in Python to solve this optimization problem:

```python
from gurobipy import *

# Create a model
m = Model("Soup_Kitchen_Optimization")

# Define variables
x1 = m.addVar(lb=0, vtype=GRB.INTEGER, name="individual_servings")
x2 = m.addVar(lb=0, vtype=GRB.INTEGER, name="family_servings")

# Set the objective function
m.setObjective(3*x1 + 10*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(250*x1 + 1200*x2 <= 50000, "soup_limit")
m.addConstr(x1 >= 3*x2, "individual_vs_family")
m.addConstr(x2 >= 10, "family_servings_min")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Individual servings: {x1.x}")
    print(f"Family servings: {x2.x}")
    print(f"Maximum profit: ${3*x1.x + 10*x2.x}")
else:
    print("No optimal solution found")
```