$\text{ If }(x-3a)(x-a-3)<0\mathrm{\forall }x\in [1,3]\text{ then exhaustive set of values of }a\text{ is }$

$\begin{array}{l}(x-3a)(x-a-3)<0\mathrm{\forall }x\in [1,3]\\ \Rightarrow (1-3a)(1-a-3)<0\end{array}$

$\Rightarrow \left(a-\frac{1}{3}\right)(a+2)<0$

$\Rightarrow -2<a<\frac{1}{3}----\left(1\right)$

$\text{ also, }(3-3a)(3-a-3)<0$

$\Rightarrow a(a-1)<0$

$\Rightarrow 0<a<1----\left(2\right)$

$\text{ Finally, }0<a<\frac{1}{3}\text{ (From (i) and (ii) common values) }$