$$\[ \mathbf{M}=\left(\begin{array}{rrr} 2 & 0 & 3 \\ 0 & -4 & -3 \\ 0 & -4 & 0 \end{array}\right) \]$$ Given that $$\(\mathbf{M}\)$$ has exactly two distinct eigenvalues $$\(\lambda_{1}\)$$ and $$\(\lambda_{2}\)$$ where $$\(\lambda_{1}< \lambda_{2}\)$$ determine a normalised eigenvector corresponding to the eigenvalue $$\(\lambda_{1}\)$$ (6)

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