The real numbers $x_1,x_2,x_3$  satisfying the equation $x^3-x^2+\beta x+\gamma =0$   are in A.P, then

The real roots of $x^3-x^2+\beta x+\gamma =0$

Given $x_1,x_2,x_3$  are in A.P

Let $x_1=a-d,x_2=a,x_3=a+d$

Sum of the roots $a-d+a+a+d=1$

$3a=1\Rightarrow a=\frac{1}{3}$

$x_1{x}_2+x_2{x}_3+x_3{x}_1=\beta$

$\left(a-d\right)a+a\left(a+d\right)\left(a+d\right)\left(a-d\right)=\beta$

$x_1{x}_2{x}_3=-\gamma \cdots \cdots \cdots \left(3\right)$

$\left(\frac{1}{3}-d\right)\frac{1}{3}+\frac{1}{3}\left(\frac{1}{3}+d\right)+\left(\frac{1}{9}-d^2\right)=\beta$

$\frac{1}{3}-d^2=\beta$

$\frac{1}{3}-d^2<\frac{1}{3}$

$\therefore \beta \in \left(-\infty ,\frac{1}{3}\right)$

From $\left(3\right)$

$\left(\frac{1}{3}-d\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}+d\right)=-\gamma$

$\left(\frac{1}{9}-d^2\right)\frac{1}{3}=-r$

$\frac{1}{27}-\frac{d^2}{3}=-\gamma$

$\frac{1}{3}\left(\frac{1}{9}-d^2\right)>\frac{1}{3}\left(-\frac{1}{9}\right)=-\frac{1}{27}$

$\therefore \gamma \in \left(-\frac{1}{27},\infty \right)$