Two of the roots of the equation $z^3 + az^2 + bz + c = 0$ are $3\text{e}^{\text{i}\left(-\frac{2}{3}\pi\right)}$ and $-2$.
Given further that $a$, $b$ and $c$ are integer constants, find the values of $a$, $b$ and $c$.

1. Apply the conjugate root theorem to identify the third root.

Since the coefficients $a$, $b$, and $c$ are integer constants (and thus real), the complex roots of the polynomial must occur in conjugate pairs.
Given the root $z_1 = 3\text{e}^{\text{i}\left(-\frac{2}{3}\pi\right)}$, the third root must be its complex conjugate:
$z_2 = 3\text{e}^{\text{i}\left(\frac{2}{3}\pi\right)}$

2. Form the quadratic factor corresponding to the complex conjugate roots.

The sum of the conjugate roots is:
$3\text{e}^{\text{i}\left(-\frac{2}{3}\pi\right)} + 3\text{e}^{\text{i}\left(\frac{2}{3}\pi\right)} = 3\left(2\cos\left(\frac{2}{3}\pi\right)\right) = 6\left(-\frac{1}{2}\right) = -3$
The product of the conjugate roots is:
$\left(3\text{e}^{\text{i}\left(-\frac{2}{3}\pi\right)}\right)\left(3\text{e}^{\text{i}\left(\frac{2}{3}\pi\right)}\right) = 9\text{e}^0 = 9$
Thus, the quadratic factor formed by these roots is $z^2 - (\text{sum})z + (\text{product})$:
$z^2 - (-3)z + 9 = z^2 + 3z + 9$

3. Multiply the quadratic factor by the linear factor of the real root to find $a, b,$ and $c$.

Since $-2$ is also a root, $(z - (-2)) = (z + 2)$ is a factor of the cubic polynomial. We can express the equation as:
$(z + 2)(z^2 + 3z + 9) = 0$
Expanding this expression:
$z(z^2 + 3z + 9) + 2(z^2 + 3z + 9) = 0$
$z^3 + 3z^2 + 9z + 2z^2 + 6z + 18 = 0$
$z^3 + 5z^2 + 15z + 18 = 0$
By comparing coefficients with $z^3 + az^2 + bz + c = 0$, we find:
$a = 5, \quad b = 15, \quad c = 18$