The smallest value of K, for which both the roots of the equation $x^2-6Kx+9(K^2-K+1)=0$  are real, distinct and have values atleast 3 is

$x^2-6Kx+9(K^2-K+1)=0$

$$\begin{gathered} \Delta >0 \\ \Rightarrow {\left(6K\right)}^2-4\left(9\right)(K^2-K+1)>0 \\ \Rightarrow 36(K-1)>0\Rightarrow K>1 \dots \dots . \left(1\right) \\ -\frac{b}{2a}>3\Rightarrow \frac{6K}{2}>3\Rightarrow K>1 \dots \dots . \left(2\right) \end{gathered}$$

$$\begin{gathered} f\left(3\right)\ge 0\Rightarrow 9-18K+9(K^2-K+1)\ge 0 \\ \Rightarrow 9(K^2-3K+2)\ge 0 \\ K\le 1\cup K\ge 2 \dots \dots . \left(3\right) \end{gathered}$$

Using (1), (2), and (3)  $K\in [2,\infty )$

$K_{\min }=2$