Calculate the annual mass of coal required, particulates produced, and CO2 produced, given the annual heat input required and the properties of the coal.

The annual heat input required ($Q_{in}$) is $5.230251 \times 10^{10} \text{ MJ/year}$.
The calorific value of coal is $29.7 \text{ MJ/kg}$.

$$M_{coal} = \frac{Q_{in}}{\text{Calorific Value}}$$ $$M_{coal} = \frac{5.230251 \times 10^{10} \text{ MJ/year}}{29.7 \text{ MJ/kg}}$$ $$M_{coal} = 1.7609936 \times 10^9 \text{ kg/year}$$ $$M_{coal} \approx 1.76 \times 10^9 \text{ kg/year}$$

To convert to metric tons ($1 \text{ ton} = 1000 \text{ kg}$):

$$M_{coal} = 1.76 \times 10^6 \text{ tons/year}$$

Particulates are primarily composed of ash. The ash content in the coal is $7.7\%$.

$$M_{particulates} = M_{coal} \times \text{Ash content}$$ $$M_{particulates} = 1.7609936 \times 10^9 \text{ kg/year} \times 0.077$$ $$M_{particulates} = 1.3569651 \times 10^8 \text{ kg/year}$$ $$M_{particulates} \approx 1.36 \times 10^8 \text{ kg/year}$$

To convert to metric tons:

$$M_{particulates} = 1.36 \times 10^5 \text{ tons/year}$$

The amount of particulates produced annually is $\boxed{\text{1.36} \times 10^8 \text{ kg/year}}$.

The carbon content in the coal is $77.0\%$.
The combustion reaction for carbon is $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$.
Molar mass of C $\approx 12.01 \text{ g/mol}$.
Molar mass of CO$_2 \approx 44.01 \text{ g/mol}$.
The mass of carbon in the coal is:

$$M_C = M_{coal} \times \text{Carbon content}$$ $$M_C = 1.7609936 \times 10^9 \text{ kg/year} \times 0.770$$ $$M_C = 1.3569651 \times 10^9 \text{ kg/year}$$

The mass of CO$_2$ produced is:

$$M_{CO_2} = M_C \times \frac{\text{Molar mass of CO}_2}{\text{Molar mass of C}}$$ $$M_{CO_2} = 1.3569651 \times 10^9 \text{ kg/year} \times \frac{44.01}{12.01}$$ $$M_{CO_2} = 4.97400 \times 10^9 \text{ kg/year}$$ $$M_{CO_2} \approx 4.97 \times 10^9 \text{ kg/year}$$

To convert to metric tons:

$$M_{CO_2} = 4.97 \times 10^6 \text{ tons/year}$$

The amount of CO$_2$ produced annually is $\boxed{\text{4.97} \times 10^9 \text{ kg/year}}$.

The sulfur content in the coal is $1.0\%$.
The combustion reaction for sulfur is $\text{S} + \text{O}_2 \rightarrow \text{SO}_2$.
Molar mass of S $\approx 32.07 \text{ g/mol}$.
Molar mass of SO$_2 \approx 64.07 \text{ g/mol}$.
The mass of sulfur in the coal is:

$$M_S = M_{coal} \times \text{Sulfur content}$$ $$M_S = 1.7609936 \times 10^9 \text{ kg/year} \times 0.010$$ $$M_S = 1.7609936 \times 10^7 \text{ kg/year}$$

The mass of SO$_2$ produced is:

$$M_{SO_2} = M_S \times \frac{\text{Molar mass of SO}_2}{\text{Molar mass of S}}$$ $$M_{SO_2} = 1.7609936 \times 10^7 \text{ kg/year} \times \frac{64.07}{32.07}$$ $$M_{SO_2} = 3.5182 \times 10^7 \text{ kg/year}$$ $$M_{SO_2} \approx 3.52 \times 10^7 \text{ kg/year}$$

To convert to metric tons:

$$M_{SO_2} = 3.52 \times 10^4 \text{ tons/year}$$

The amount of SO$_2$ produced annually is $\boxed{\text{3.52} \times 10^7 \text{ kg/year}}$.