If both the roots of the equation $x^2-6ax+2-2a+9a^2=0$ exceed 3, then

We can write the given equation

$(x-3a{)}^2=2a-2$

Note that a ≥ 1 and

$x=3a\pm \sqrt{2a-2}$

Both the roots will exceed 3 if smaller of the two roots exceed 3, that is, if

$3a-\sqrt{2a-2}>3$

$\Rightarrow 3(a-1)>\sqrt{2}\sqrt{a-1}$

$\therefore a>1$and $\sqrt{a-1}>\sqrt{2}/3$

$\Rightarrow a>1+\frac{2}{9}=\frac{11}{9}$