1. The problem is to find the domain of the function $f(x) = \sqrt{2 - 3x + 3}$.\n\n2. The domain of a square root function requires the expression inside the root to be greater than or equal to zero because the square root of a negative number is not a real number.\n\n3. Set the inside of the square root greater than or equal to zero:\n$$2 - 3x + 3 \geq 0$$\n\n4. Simplify the inequality:\n$$5 - 3x \geq 0$$\n\n5. Solve for $x$:\n$$5 \geq 3x$$\n$$\cancel{3} \cdot \frac{5}{\cancel{3}} \geq \cancel{3} \cdot x$$\n$$\frac{5}{3} \geq x$$\n\n6. Rewrite the inequality in interval notation. Since $x$ must be less than or equal to $\frac{5}{3}$, the domain is:\n$$(-\infty, \frac{5}{3}]$$\n\nTherefore, the domain of $f(x)$ is all real numbers less than or equal to $\frac{5}{3}$.