A person invested $\$ 6100$ for 1 year, part at $5 \%$, part at $11 \%$, and the remainder at $13 \%$. The total annual income from these investments was $\$ 665$. The amount of money invested at $13 \%$ was $\$ 500$ more than the amounts invested at $5 \%$ and $11 \%$ combined. Find the amount invested at each rate.

The person invested $\$ \square$ at $5 \%, \$ \square$ at $11 \%$, and $\$ \square$ at $13 \%$.

Step 1 ：Let's denote the amount invested at 5% as \(x\), the amount invested at 11% as \(y\), and the amount invested at 13% as \(z\).

Step 2 ：From the problem, we know that the total amount invested is $6100, so we have the equation \(x + y + z = 6100\).

Step 3 ：We also know that the total annual income from these investments was $665, which gives us the equation \(0.05x + 0.11y + 0.13z = 665\).

Step 4 ：Finally, we know that the amount of money invested at 13% was $500 more than the amounts invested at 5% and 11% combined, so we have the equation \(z = x + y + 500\).

Step 5 ：Solving this system of equations, we find that \(x = 1200\), \(y = 1600\), and \(z = 3300\).

Step 6 ：So, the final answer is: The person invested \(\boxed{\$ 1200}\) at 5%, \(\boxed{\$ 1600}\) at 11%, and \(\boxed{\$ 3300}\) at 13%.