$a + b + c = 2022$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{2022}}$$S = \frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}} = ?$

We have, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{2022}} = \frac{1}{{a + b + c}}$

$\Rightarrow \frac{1}{a} + \frac{1}{b} + \left( {\frac{1}{c} - \frac{1}{{a + b + c}}} \right) = 0$

$\Rightarrow \frac{{a + b}}{{ab}} + \frac{{a + b}}{{(a + b + c)c}} = 0$

$\Rightarrow (a + b)\frac{{(ab + bc + ca + {c^2})}}{{abc(a + b + c)}} = 0$

$\Rightarrow (a + b)\frac{{(b + c)(c + a)}}{{abc(a + b + c)}} = 0$

$\Rightarrow (a + b)(b + c)(c + a) = 0$

$a + b = 0$ Or $b + c = 0$ Or $c + a = 0$

If $a + b = 0$, $a + b + c = 0 + c = c = 2022$

$S = \frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}} = \frac{1}{{{a^{2023}}}} + \frac{1}{{{{( - a)}^{2023}}}} + \frac{1}{{{{2022}^{2023}}}} = \frac{1}{{{{2022}^{2023}}}}$

Due to symmetry, other cases would yield the same final answer.