The correct answer is $\boxed{\text{u} = -\frac{2}{5}{\text{e}}_1 – 3{\text{e}}_2 + \frac{11}{5}{\text{e}}_3}$.
An orthogonal basis is a set of vectors that are mutually orthogonal, meaning that the dot product of any two vectors in the set is zero. In three dimensions, an orthogonal basis can be used to express any vector as a linear combination of the basis vectors.
The given vectors $\text{e}_1 = (1, 0, 2), \text{e}_2 = (0, 1, 0)$, and $\text{e}_3 = (-2, 0, 1)$ form an orthogonal basis of $\mathbb{R}^3$. Therefore, any vector $\text{u} = (u_1, u_2, u_3)$ in $\mathbb{R}^3$ can be expressed as
$$\text{u} = u_1\text{e}_1 + u_2\text{e}_2 + u_3\text{e}_3.$$
In this case, we are given that $\text{u} = (4, 3, -3)$. Substituting this into the equation above, we get
$$(4, 3, -3) = u_1\text{e}_1 + u_2\text{e}_2 + u_3\text{e}_3.$$
To solve for $u_1, u_2,$ and $u_3$, we can take the dot product of both sides of the equation with each of the basis vectors. This gives us the following system of equations:
$$4 = u_1 + 0u_2 + 2u_3$$
$$3 = 0u_1 + 3u_2 + 0u_3$$
$$-3 = -2u_1 + 0u_2 + 1u_3.$$
Solving this system of equations, we get $u_1 = -\frac{2}{5}, u_2 = 3,$ and $u_3 = \frac{11}{5}$. Therefore, the vector $\text{u}$ can be expressed as
$$\text{u} = -\frac{2}{5}{\text{e}}_1 + 3{\text{e}}_2 + \frac{11}{5}{\text{e}}_3.$$