Find the equation of the function f(x) = -x3 + ax2 + bx + c given that it has a point of inflection at (0, 27) and f(-3) = 0.

A point of inflection occurs where the second derivative $f''(x)$ is equal to zero.
First, find the first derivative $f'(x)$:

$$f'(x) = -3x^2 + 2ax + b$$

Next, find the second derivative $f''(x)$:

$$f''(x) = -6x + 2a$$

The point of inflection is at $(0; 27)$, so $f''(0) = 0$.

$$-6(0) + 2a = 0$$ $$2a = 0$$ $$a = 0$$

Also, the point $(0; 27)$ lies on the function $f(x)$, so $f(0) = 27$.

$$f(0) = -(0)^3 + a(0)^2 + b(0) + c = 27$$ $$c = 27$$

Now we know $a=0$ and $c=27$. Substitute these values back into $f(x)$:

$$f(x) = -x^3 + (0)x^2 + bx + 27$$ $$f(x) = -x^3 + bx + 27$$

Substitute $x=-3$ into the updated $f(x)$:

$$f(-3) = -(-3)^3 + b(-3) + 27 = 0$$ $$-(-27) - 3b + 27 = 0$$ $$27 - 3b + 27 = 0$$ $$54 - 3b = 0$$ $$3b = 54$$ $$b = \frac{54}{3}$$ $$b = 18$$

We have found $a=0$, $b=18$, and $c=27$. Substitute these values into the original form of $f(x)$:

$$f(x) = -x^3 + (0)x^2 + (18)x + 27$$ $$f(x) = \boxed{-x^3 + 18x + 27}$$