If the roots of the equation $x^3-12x^2+39x-28=0$  are in A.P., then common difference is

$$\begin{gathered} S_1=\alpha -\beta +\alpha +\alpha +\beta =12 \\ 3\alpha =12 \\ \therefore \alpha =4 \\ \alpha ({\alpha }^2-{\beta }^2)=28 \end{gathered}$$

Take $4(16-{\beta }^2)=28\Rightarrow {\beta }^2=9\Rightarrow \beta =\pm 3$

Take $\alpha =4,\beta =3\Rightarrow$ the roots are $1,4,7$

$\Rightarrow$common difference is 3