1. Identify the reaction and heat change

   The equilibrium is: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g), \qquad \Delta H^\circ = -198\text{ kJ}$

   Since $\Delta H^\circ$ is negative, the forward reaction is exothermic.

2. Effect of temperature using Le Chatelier's principle

   For an exothermic reaction, heat may be treated as a product: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) + \text{heat}$

   If temperature is lowered, the equilibrium shifts in the direction that produces heat, i.e. forward direction.

   So, low temperature favors the forward reaction.

3. Effect of pressure using Le Chatelier's principle

   Count moles of gaseous reactants and products:

   • Reactant side: $2 + 1 = 3$ moles of gas
   • Product side: $2$ moles of gas

   Increasing pressure shifts equilibrium toward the side with fewer moles of gas.

   Since the product side has fewer moles, high pressure favors the forward reaction.

4. Combine both conditions

   • Forward reaction is favored by lower temperature
   • Forward reaction is favored by higher pressure

   Therefore, the favorable condition is: $\boxed{\text{lowering temperature and increasing pressure}}$

5. Evaluate options

   • A: increasing temperature as well as pressure $\rightarrow$ temperature effect is unfavorable
   • B: lowering the temperature and increasing the pressure $\rightarrow$ both effects favor forward reaction
   • C: any value of temperature and pressure $\rightarrow$ incorrect
   • D: lowering temperature as well as pressure $\rightarrow$ pressure effect is unfavorable

Hence, the correct option is B.