jmc

Note that $z^2+z+1=0$ and $z^3=1.$ Also, note that $s_2=z{s}_1$ and $s_3=z^2{s}_1.$

The sum of the coefficients of $g(x)$ is $$\begin{array}{rl}g(1)& =(1-s_1)(1-s_2)(1-s_3)\\ & =(1-s_1)(1-s_{1}z)(1-s_1{z}^2)\\ & =1-(1+z+z^2)s_1+(z+z^2+z^3)s_1^2-z^3{s}_1^3\\ & =1-s_1^3.\end{array}$$We have that $$s_1^3=r_1^3+r_2^3+r_3^3+3r_1^2{r}_{2}z+3r_1^2{r}_3{z}^2+3r_2^2{r}_{3}z+3r_2^2{r}_1{z}^2+3r_3^2{r}_{1}z+3r_3^2{r}_2{z}^2+6r_1{r}_2{r}_3.$$Note that $r_1,$ $r_2,$ and $r_3$ are all real, and the real part of both $z$ and $z^2$ are $-\frac{1}{2},$ so the real part of $s_1^3$ is $$\begin{array}{rl}& r_1^3+r_2^3+r_3^3-\frac{3}{2}(r_1^2{r}_2+r_1{r}_2^2+r_1^2{r}_3+r_1{r}_3^2+r_2^2{r}_3+r_2{r}_3^2)+6r_1{r}_2{r}_3\\ & =(r_1+r_2+r_3{)}^3-\frac{9}{2}(r_1+r_2+r_3)(r_1{r}_2+r_1{r}_3+r_2{r}_3)+\frac{27}{2}{r}_1{r}_2{r}_3\\ & =3^3-\frac{9}{2}(3)(-4)+\frac{27}{2}(-4)=27.\end{array}$$Hence, the real part of the sum of the coefficients of $g(x)$ is $1-27=\boxed{-26}.$