The least positive integer value of a,  so that any solutions of the inequality  $\frac{{\log }_3(x^2-3x+7)}{{\log }_3(3x+2)}<1$, is also a solution  of the inequality  $x^2+(5-2a)x-10a\le 0$  is ____

$\frac{{\log }_3(x^2-3x+7)}{{\log }_3(3x+2)}<1\Rightarrow x\in (-\frac{2}{3},-\frac{1}{3})\cup (1,5)$

For every solution of the inequality, to be the solution of the inequality
$x^2+(5-2a)x-10a\le 0,$  the zeroes of  $x_1,x_2$ must satisfy  $x_1\le -\frac{2}{3}\&x_2\ge 5$
$\Rightarrow f(-\frac{2}{3})\le 0\&f\left(5\right)\le 0 \Rightarrow {(-\frac{2}{3})}^2+(5-2a)(-\frac{2}{3})-10a\le 0\&5^2+(5-2a)5-10a\le 0$

$\Rightarrow a\in [\frac{5}{2},\infty )$