Let, the point $\left(0,a\right)$ lie inside the circle whose end-points of the diameter are $\left(-\mathrm{2,0}\right)$ and $\left(\mathrm{1,1}\right)$, but does not lie on this diameter
$\Rightarrow \left(0,a\right)$ lie inside the circle $\left(x+2\right)\left(x-1\right)+y\left(y-1\right)=0$
$\Rightarrow \left(0+2\right)\left(0-1\right)+a\left(a-1\right)<0$
$\Rightarrow a^2-a-2<0\Rightarrow \left(a-2\right)\left(a+1\right)<0$
$\Rightarrow a\in \left(-1,2\right)$
But, $a\ne \frac{2}{3}$ because at $a=\frac{2}{3}$, points $\left(0,\frac{2}{3}\right),\left(-2,0\right)$ and $\left(1,1\right)$ are collinear
$\Rightarrow a\in \left(-1,2\right)-\left\{\frac{2}{3}\right\}$ 

Hence, the sum of all possible integral values of $‘a’$ is $1$