Given,

$${{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0$$

$${x^2} + 3x + 5$$ is a quadratic equation

$$a = 1 > 0$$ and $$D = {( - 3)^2} - 4\,.\,1\,.\,5 = - 11 < 0$$

$$\therefore$$ $${x^2} + 3x + 5 > 0$$ (always)

So, we can ignore this quadratic term

$${{(x + 2)} \over { - 2 + 3x - {x^2}}} \ge 0$$

$$ \Rightarrow {{x + 2} \over { - ({x^2} - 3x + 2)}} \ge 0$$

$$ \Rightarrow {{x + 2} \over {{x^2} - 3x + 2}} \le 0$$

$$ \Rightarrow {{x + 2} \over {{x^2} - 2x - x + 2}} \le 0$$

$$ \Rightarrow {{x + 2} \over {(x - 1)(x - 2)}} \le 0$$

$$\therefore$$ $$x \in ( - \alpha , - 2] \cup (1,2)$$

$$\therefore$$ $${S_1} = ( - \alpha , - 2] \cup (1,2)$$

Now,

$${3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0$$

$$ \Rightarrow {({3^x})^2} - 3\,.\,{3^x} - {3^2}\,.\,{3^x} + 27 \le 0$$

Let $${3^x} = t$$

$$ \Rightarrow {t^2} - 3\,.\,t - {3^2}\,.\,t + 27 \le 0$$

$$ \Rightarrow t(t - 3) - 9(t - 3) \le 0$$

$$ \Rightarrow (t - 3)(t - 9) \le 0$$

$$\therefore$$ $$3 \le t \le 9$$

$$ \Rightarrow {3^1} \le {3^x} \le {3^2}$$

$$ \Rightarrow 1 \le x \le 2$$

$$\therefore$$ $$x \in [1,2]$$

$$\therefore$$ $${S_2} = [1,2]$$

$$\therefore$$ $${S_1} \cup {S_2} = ( - \alpha ,2] \cup (1,2) \cup [1,2]$$

$$ = ( - \alpha ,2] \cup [1,2]$$