The value of ‘a’ for which the quadratic expression $a{x}^2+|2a-3|x-6$  is positive for exactly  three integral values of x is 

Since expression is positive only for three integral values of x, therefore parabola
$y=a{x}^2+|2a-3|x-6$ will open downward
$\Rightarrow$$a<0$  and  $y=a{x}^2+(3-2a)x-6ory=(ax+3)(x-2)$
 $\Rightarrow 5<-\frac{3}{a}\le 6$
 $\Rightarrow -\frac{3}{5}<a\le -\frac{1}{2}.$