According to Hess's Law of Constant Heat Summation, if a reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided .

By adding the three provided thermochemical equations, we can derive the target equation for the combustion of carbon:

1. $H_{2}O(g) + C(s) ightarrow CO(g) + H_{2}(g) \quad (\Delta H_{1} = 131 \text{ kJ})$
2. $CO(g) + \frac{1}{2}O_{2}(g) ightarrow CO_{2}(g) \quad (\Delta H_{2} = -282 \text{ kJ})$
3. $H_{2}(g) + \frac{1}{2}O_{2}(g) ightarrow H_{2}O(g) \quad (\Delta H_{3} = -242 \text{ kJ})$

Summing these equations: $(H_{2}O(g) + C(s) + CO(g) + \frac{1}{2}O_{2}(g) + H_{2}(g) + \frac{1}{2}O_{2}(g)) ightarrow (CO(g) + H_{2}(g) + CO_{2}(g) + H_{2}O(g))$

Cancelling common terms on both sides ($H_{2}O(g), CO(g), H_{2}(g)$) yields the net reaction: $C(s) + O_{2}(g) ightarrow CO_{2}(g)$

The total enthalpy change ($X$) is the sum of the individual enthalpy changes: $X = \Delta H_{1} + \Delta H_{2} + \Delta H_{3}$ $X = 131 \text{ kJ} + (-282 \text{ kJ}) + (-242 \text{ kJ})$ $X = 131 \text{ kJ} - 524 \text{ kJ} = -393 \text{ kJ}$.

Thus, the value of $X$ is –393 kJ, matching Option A.