Given equation $\mathrm{x}^3-9 \mathrm{x}^2+\mathrm{k}=0$ where $\mathrm{k} \in \mathrm{R}-\{0\}$.
Here, $\alpha, \beta \& 2 \beta$ are roots.


$\begin{aligned} & \alpha \beta+\beta(2 \beta)+2 \beta(\alpha)=0 \\ & \alpha \beta+2 \beta^2+2 \beta \alpha=0\end{aligned}$

From (i),
$$
\alpha+3 \beta=9
$$
From (ii)
$$
\begin{aligned}
& 2 \beta^2+3 \alpha \beta=0 \\
& \beta(2 \beta+3 \alpha)=0 \\
& \beta=0,2 \beta+3 \alpha=0 \\
& 2 \beta=-3 \alpha \\
& \alpha=\frac{-2 \beta}{3}
\end{aligned}
$$
Put value of $\alpha$ in equation $\alpha+3 \beta=9$.
$$
\begin{aligned}
& \Rightarrow \frac{-2 \beta}{3}+3 \beta=9 \\
& \frac{-2 \beta+9 \beta}{3}=9 \\
& 7 \beta=27 \\
& \beta=\frac{27}{7}
\end{aligned}
$$
Now, $14 \beta=14 \times \frac{27}{7}=54$
So, option (d) is correct.