1. The problem asks to find the residue of the function $$f(z) = \frac{z+2}{(z+1)(z-2)(z-3)}$$ at the point $$z_0 = 2$$. 2. The residue at a simple pole $$z_0$$ of a function $$f(z) = \frac{g(z)}{h(z)}$$ where $$h(z_0) = 0$$ and $$h'(z_0) \neq 0$$ is given by: $$\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z)$$ 3. Here, $$z_0 = 2$$ is a simple pole because the denominator has a factor $$z-2$$. 4. Calculate the residue: $$\text{Res}(f, 2) = \lim_{z \to 2} (z - 2) \frac{z+2}{(z+1)(z-2)(z-3)} = \lim_{z \to 2} \frac{z+2}{(z+1)(z-3)}$$ 5. Substitute $$z = 2$$: $$\text{Res}(f, 2) = \frac{2+2}{(2+1)(2-3)} = \frac{4}{3 \times (-1)} = -\frac{4}{3}$$ 6. Therefore, the residue at $$z=2$$ is $$-\frac{4}{3}$$. Final answer: c. $$-\frac{4}{3}$$