Determine the sum of all numbers for which the following expression is not defined. $$\left(\frac{x-2}{x^2-9}+1\right):\frac{1-x}{x+2} + \frac{5}{3x-x^2}$$ Camille solved the task in the following steps:
(1) For the expression in parentheses, she wrote the existence condition and determined for which values of $x$ this expression is not defined: $$\begin{aligned}x^2 - 9\neq0 &\Rightarrow (x - 3)\cdot(x + 3)\neq0 \Rightarrow \cr&\Rightarrow(x - 3\neq0 \land x + 3 \neq 0)\Rightarrow \cr &\Rightarrow (x\neq3 \land x \neq -3)\end{aligned}$$ (2) Then, she wrote the condition that must be satisfied for the fraction $\frac{1-x}{x+2}$ to be defined: $$x + 2\neq0 \Rightarrow x \neq -2$$
(3) Further, Camille wrote the existence condition for the fraction $\frac{5}{3x-x^2}$ and determined for which values of $x$ this fraction is not defined: $$\begin{aligned} 3x-x^2\neq0 &\Rightarrow x\cdot(3-x)\neq0 \Rightarrow\cr &\Rightarrow (x\neq0 \land 3 - x\neq 0)\Rightarrow\cr & \Rightarrow (x \neq0 \land x \neq 3)\end{aligned}$$ (4) Finally, Camille claimed that the given expression is not defined for $$x\in\left\{-3;-2;0;3\right\}$$ and determined the sum of these numbers: $(-3) + (-2) + 0 + 3 = -2$.
There is a mistake in Camille's solution. Where did Camille make a mistake in her procedure?
The mistake is in step (1). Camille did not correctly determine all the conditions for the expression in parentheses.
The mistake is in step (2). Camille did not correctly determine all the conditions for the fraction $\frac{1-x}{x+2}$ to be defined.
The mistake is in step (3). Camille did not correctly determine all the conditions that the fraction $\frac{5}{3x-x^2}$ must satisfy to make the given expression defined.
The mistake is in step (4). Camille made the wrong conclusion since she missed one condition, and therefore, the sum of the numbers for which the given expression is not defined is not determined correctly.
Camille did not realize that the fraction $\frac{1-x}{x+2}$ is also the denominator in the expression: $$\frac{\frac{x-2}{x^2-9}+1}{\frac{1-x}{x+2}}$$ This means that the fraction $\frac{1-x}{x+2}$ must also satisfy the condition: $$\frac{1-x}{x+2}\neq 0 \Rightarrow 1 - x \neq 0 \Rightarrow x\neq1$$ The given expression is not defined for $$x\in\left\{-3;-2;0;1;3\right\}$$ and the sum of these numbers is: $$(-3) + (-2) + 0 + 1 + 3 = -1.$$