To solve the given optimization problem using Gurobi, we first need to understand and possibly simplify or clarify the constraints provided. The objective function to be maximized is:

\[8.21 \times \text{milligrams of vitamin C} + 1.87 \times \text{milligrams of vitamin B12}\]

The constraints are as follows:
- The muscle growth index contribution from milligrams of vitamin C and vitamin B12 should be at least 47 and at most 87.
- The digestive support index contribution from both vitamins should be at least 22 and at most 52.
- There's a specific linear constraint involving the quantities of both vitamins: \(6 \times \text{milligrams of vitamin C} - 3 \times \text{milligrams of vitamin B12} \geq 0\).

Given these conditions, we will formulate the problem in Gurobi. We'll define two decision variables: `vit_c` for milligrams of vitamin C and `vit_b12` for milligrams of vitamin B12.

```python
from gurobipy import *

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

# Define the decision variables
vit_c = m.addVar(lb=0, name="milligrams_of_vitamin_C", vtype=GRB.CONTINUOUS)
vit_b12 = m.addVar(lb=0, name="milligrams_of_vitamin_B12", vtype=GRB.CONTINUOUS)

# Define the objective function
m.setObjective(8.21 * vit_c + 1.87 * vit_b12, GRB.MAXIMIZE)

# Add constraints
# Muscle growth index constraint: at least 47 and at most 87
m.addConstr(12.75 * vit_c + 7.81 * vit_b12 >= 47, name="muscle_growth_min")
m.addConstr(12.75 * vit_c + 7.81 * vit_b12 <= 87, name="muscle_growth_max")

# Digestive support index constraint: at least 22 and at most 52
m.addConstr(0.69 * vit_c + 1.13 * vit_b12 >= 22, name="digestive_support_min")
m.addConstr(0.69 * vit_c + 1.13 * vit_b12 <= 52, name="digestive_support_max")

# Additional linear constraint
m.addConstr(6 * vit_c - 3 * vit_b12 >= 0, name="additional_constraint")

# Solve the model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Milligrams of Vitamin C: {vit_c.x}")
    print(f"Milligrams of Vitamin B12: {vit_b12.x}")
    print(f"Objective Function Value: {m.objVal}")
else:
    print("No optimal solution found")
```
```python
```