If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-2 x^2+3 x-4=0$, then
$\begin{aligned} & \alpha+\beta+\gamma=\frac{2}{1}=2 \\ & \alpha \beta+\beta \gamma+\gamma \alpha=3 \\ & \text { and } \\ & \alpha \beta \gamma=4 \\ & \end{aligned}$
We know that
$\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2$
$\begin{aligned} & =(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2(\alpha \beta \gamma)(\alpha+\beta+\gamma) \\ & =(3)^2-2(4)(2) \\ & =9-16 \\ & =-7\end{aligned}$