Given equation
$x^4-3 x^2+6 x-12=0$ ...(i)
On comparing with $a x^4+b x^3+c x^2+d x+e=0$
$a=1, b=0, c=-3, d=6, e=-12$
$\because \alpha, \beta, \gamma, \delta$ are the zeroes of polynomial equation.
$\therefore \quad \alpha+\beta+\gamma+\delta=\frac{-b}{a}=0$
$\Rightarrow \quad \alpha+\beta+\gamma=-\delta, \alpha+\delta+\gamma=-\beta$
$\alpha+\beta+\delta=-\gamma, \delta+\beta+\gamma=-\alpha$
$\begin{gathered}\Sigma \alpha \beta=\frac{c}{a}=-3 \\ \Sigma \alpha \beta \gamma=\frac{-d}{a}=-6 \\ \alpha \beta \gamma \delta=\frac{e}{a}=-12\end{gathered}$
Consider, $\frac{\alpha+\beta+\gamma}{\delta^2}+\frac{\alpha+\delta+\gamma}{\beta^2}$ $+\frac{\alpha+\beta+\delta}{\gamma^2}+\frac{\delta+\beta+\gamma}{\alpha^2}$
$\begin{aligned} & =\frac{-\delta}{\delta^2}+\frac{-\beta}{\beta^2}+\frac{-\gamma}{\gamma^2}+\frac{-\alpha}{\alpha^2} \\ & =-\frac{1}{\delta}-\frac{1}{\beta}-\frac{1}{\gamma}-\frac{1}{\alpha} \\ & =-\left(\frac{\alpha \beta \gamma+\alpha \gamma \delta+\alpha \beta \delta+\beta \gamma \delta}{\alpha \beta \gamma \delta}\right) \\ & =-\left(\frac{\Sigma \alpha \beta \gamma}{\alpha \beta \gamma \delta}\right)=-\left(\frac{-6}{-12}\right)=\frac{-1}{2}\end{aligned}$