The given function is:

$f(x) = \sqrt{(3 - x)(x + 2)} + \log_{x+3} |x|$

Steps to find the domain of $f(x)$:

1. Square Root Condition: The expression inside the square root, $(3 - x)(x + 2)$, must be non-negative: $(3 - x)(x + 2) \geq 0$

   We solve this inequality by finding the critical points ($x = -2$ and $x = 3$) and testing intervals:

   • For $x < -2$: Both factors are negative, so their product is positive ($3 - x > 0$ and $x + 2 < 0$).
   • For $-2 < x < 3$: One factor is positive, and one is negative, so their product is negative.
   • For $x > 3$: Both factors are positive, so their product is positive.

   Therefore, the solution for the inequality is: $x \in (-\infty, -2] \cup [3, \infty)$

2. Logarithmic Condition: The base of the logarithm, $x + 3$, must be positive ($x > -3$), and the argument, $|x|$, must also be positive ($x \neq 0$).

   Combining these conditions: $x > -3 \quad \text{and} \quad x \neq 0$

   This means: $x \in (-3, 0) \cup (0, \infty)$

3. Combining Conditions: The domain of the function must satisfy both conditions simultaneously: $x \in (-\infty, -2] \cup [3, \infty) \quad \text{and} \quad x \in (-3, 0) \cup (0, \infty)$

   The intersection of these sets is: $x \in (-2, 0) \cup (0, 3]$

Final Answer:

The domain of $f(x)$ is: $\boxed{(-2, 0) \cup (0, 3]}$

This corresponds to option (в) in the question.

Would you like further clarification or a detailed graph of the domain? Here are five questions to expand your understanding:

1. Why must the square root require non-negative values inside?
2. How does the base of the logarithm affect the domain?
3. Can you explain how to find the intersection of intervals?
4. How would the domain change if the absolute value in $\log$ were not present?
5. Can we solve a similar problem involving cube roots instead of square roots?

Tip: Always analyze each function component separately before combining conditions for the domain!