If one root of ${\mathrm{x}}^3 – 12{\mathrm{x}}^2 + \mathrm{kx} – 18 = 0$ is thrice the sum of remaining two roots then k =

$\begin{array}{l}Let\alpha ,\beta ,\gamma arerootsof{x}^3-12x^2+kx-18=0\\ and\alpha =3(\beta +\gamma )\Rightarrow \beta +\gamma =\frac{\alpha }{3}then\\ \therefore \alpha +\beta +\gamma =12\\ \Rightarrow \alpha +\frac{\alpha }{3}=12\Rightarrow \frac{4\alpha }{3}=12\Rightarrow \alpha =9\\ \because \alpha isroot\Rightarrow {\left(9\right)}^3-12{\left(9\right)}^2+k\left(9\right)-18=0\\ \Rightarrow 9^2-12\left(9\right)+k-2=0\\ \Rightarrow 81-108+k-2=0\\ \Rightarrow k-29=0\\ \Rightarrow k=29\end{array}$