## Step 1: Define the symbolic representation of the problem
Let's denote the number of packets of cat food as $x_1$ and the number of cans of tuna as $x_2$. The objective is to minimize the total cost, which is $12x_1 + 7x_2$. The constraints are based on the nutritional requirements: $5x_1 + 7x_2 \geq 700$ (carbohydrates), $15x_1 + 12x_2 \geq 1100$ (vitamins), and $12x_1 + 15x_2 \geq 900$ (protein). Also, $x_1 \geq 0$ and $x_2 \geq 0$ since the quantities cannot be negative.

## Step 2: Express the problem in the required symbolic format
The symbolic variables are:
- $x_1$ for packets of cat food
- $x_2$ for cans of tuna

The objective function to minimize is: $12x_1 + 7x_2$

The constraints are:
- $5x_1 + 7x_2 \geq 700$
- $15x_1 + 12x_2 \geq 1100$
- $12x_1 + 15x_2 \geq 900$
- $x_1 \geq 0$
- $x_2 \geq 0$

In the required format:
```json
{
'sym_variables': [('x1', 'packets of cat food'), ('x2', 'cans of tuna')],
'objective_function': '12*x1 + 7*x2',
'constraints': [
    '5*x1 + 7*x2 >= 700',
    '15*x1 + 12*x2 >= 1100',
    '12*x1 + 15*x2 >= 900',
    'x1 >= 0',
    'x2 >= 0'
]
}
```

## 3: Convert the problem into Gurobi code
To solve this linear programming problem using Gurobi, we will use the Gurobi Python API.

```python
import gurobi

def solve_cat_food_problem():
    # Create a new model
    model = gurobi.Model()

    # Define the variables
    x1 = model.addVar(name="cat_food", lb=0, vtype=gurobi.GRB.CONTINUOUS)
    x2 = model.addVar(name="tuna", lb=0, vtype=gurobi.GRB.CONTINUOUS)

    # Define the objective function
    model.setObjective(12 * x1 + 7 * x2, gurobi.GRB.MINIMIZE)

    # Add constraints
    model.addConstr(5 * x1 + 7 * x2 >= 700, name="carbohydrates")
    model.addConstr(15 * x1 + 12 * x2 >= 1100, name="vitamins")
    model.addConstr(12 * x1 + 15 * x2 >= 900, name="protein")

    # Optimize the model
    model.optimize()

    # Print the solution
    if model.status == gurobi.GRB.OPTIMAL:
        print("Optimal solution found.")
        print(f"Packets of cat food: {x1.varValue}")
        print(f"Cans of tuna: {x2.varValue}")
        print(f"Total cost: {model.objVal}")
    else:
        print("No optimal solution found.")

solve_cat_food_problem()
```