Let $\mathrm{x}=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$ and $\mathrm{A}=\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]$ . For $\mathrm{k}\in \mathrm{\mathbb{N}}\text{, if }\mathrm{X}\text{'}{\mathrm{A}}^{\mathrm{k}}\mathrm{X}=\left[33\right]$ then K is equal to :

$\begin{array}{l}\mathrm{X}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right];\mathrm{A}=\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]\\ {\mathrm{X}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{K}}\mathrm{X}=33\\ \left[\begin{array}{lll}1& 1& 1\end{array}\right]{\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]}^{\mathrm{k}}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=33\\ \left[\begin{array}{lll}1& 1& 1\end{array}\right]{\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]}^{\mathrm{k}}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=33\end{array}$
$\begin{array}{l}\text{ As }{\mathrm{A}}^2=\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]=\left[\begin{array}{lll}1& 0& 6\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ {\mathrm{A}}^4=\left[\begin{array}{lll}1& 0& 6\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{lll}1& 0& 6\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 12\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ {\mathrm{A}}^8=\left[\begin{array}{ccc}1& 0& 24\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ {\mathrm{A}}^{10}=\left[\begin{array}{lll}1& 0& 6\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 24\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 30\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ \end{array}$
for $\mathrm{K}\to \text{ Even }{\mathrm{A}}^{\mathrm{k}}=\left[\begin{array}{ccc}1& 0& 3\mathrm{K}\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\begin{array}{l}{\mathrm{X}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{K}}\mathrm{X}=\left[33\right]\\ \left[\begin{array}{lll}1& 1& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 3\mathrm{K}\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]\\ =\left[\begin{array}{lll}1& 1& 3\mathrm{K}+1\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=[3\mathrm{K}+3]\\ \therefore 3\mathrm{K}+3=33\\ \boxed{\therefore \mathrm{K}=10}\\ \text{ If }\mathrm{K}\text{ is odd }\\ {\mathrm{X}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{K}}\mathrm{X}=33\\ {\mathrm{X}}^{\mathrm{T}}{\mathrm{AA}}^{\mathrm{K}-1}\mathrm{X}=33\\ \left[\begin{array}{lll}1& 1& 1\end{array}\right]\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]\left[\begin{array}{ccc}1& 0& 3\mathrm{K}-3\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=33\\ [-3\mathrm{K}+13]=\left[33\right]\\ \end{array}$
$\mathrm{k}=\frac{20}{3}$(not possible)