[tex]\text{We know}\ \sqrt{a}\ \text{is exist for}\ a\geq0.\\\\\text{Therefore the domain of}\ f(x)=3\sqrt{x-2}\ \text{is:}\\\\x-2\geq0\qquad\text{add 2 to both sides}\\\\\boxed{x\geq2\to x\in[2,\ \infty)}[/tex]
