1. **Problem statement:** For the matrix $$A = \begin{pmatrix} 3 & -2 & 2 \\ 0 & -2 & 1 \\ 0 & 0 & 1 \end{pmatrix},$$ verify which of the vectors $$\mathbf{v}_a = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix}, \mathbf{v}_b = \begin{pmatrix} -3 \\ 2 \\ -1 \end{pmatrix}, \mathbf{v}_c = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}, \mathbf{v}_d = \begin{pmatrix} 2 \\ 5 \\ 0 \end{pmatrix}$$ is an eigenvector corresponding to the eigenvalue $$\lambda = -2$$. 2. **Recall the eigenvector definition:** A vector $$\mathbf{v}$$ is an eigenvector of matrix $$A$$ with eigenvalue $$\lambda$$ if and only if $$A\mathbf{v} = \lambda \mathbf{v}.$$ 3. **Check each vector:** - For $$\mathbf{v}_a$$: $$A\mathbf{v}_a = \begin{pmatrix}3 & -2 & 2 \\ 0 & -2 & 1 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}3 \\ -2 \\ 1\end{pmatrix} = \begin{pmatrix}(3)(3) + (-2)(-2) + 2(1) \\ 0(3) + (-2)(-2) + 1(1) \\ 0(3) + 0(-2) + 1(1)\end{pmatrix} = \begin{pmatrix}9 + 4 + 2 \\ 0 + 4 + 1 \\ 0 + 0 + 1\end{pmatrix} = \begin{pmatrix}15 \\ 5 \\ 1\end{pmatrix}.$$ Compare with $$\lambda \mathbf{v}_a = -2 \begin{pmatrix}3 \\ -2 \\ 1\end{pmatrix} = \begin{pmatrix}-6 \\ 4 \\ -2\end{pmatrix}$$. Since $$A\mathbf{v}_a \neq \lambda \mathbf{v}_a$$, $$\mathbf{v}_a$$ is not an eigenvector. - For $$\mathbf{v}_b$$: $$A\mathbf{v}_b = \begin{pmatrix}3 & -2 & 2 \\ 0 & -2 & 1 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}-3 \\ 2 \\ -1\end{pmatrix} = \begin{pmatrix}(3)(-3) + (-2)(2) + 2(-1) \\ 0(-3) + (-2)(2) + 1(-1) \\ 0(-3) + 0(2) + 1(-1)\end{pmatrix} = \begin{pmatrix}-9 -4 -2 \\ 0 -4 -1 \\ 0 + 0 -1\end{pmatrix} = \begin{pmatrix}-15 \\ -5 \\ -1\end{pmatrix}.$$ Multiply $$\lambda \mathbf{v}_b = -2 \begin{pmatrix}-3 \\ 2 \\ -1\end{pmatrix} = \begin{pmatrix}6 \\ -4 \\ 2\end{pmatrix}$$. Because $$A\mathbf{v}_b \neq \lambda \mathbf{v}_b$$, $$\mathbf{v}_b$$ is not an eigenvector. - For $$\mathbf{v}_c$$: $$A\mathbf{v}_c = \begin{pmatrix}3 & -2 & 2 \\ 0 & -2 & 1 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}1 \\ -2 \\ 3\end{pmatrix} = \begin{pmatrix}(3)(1) + (-2)(-2) + 2(3) \\ 0(1) + (-2)(-2) + 1(3) \\ 0(1) + 0(-2) + 1(3)\end{pmatrix} = \begin{pmatrix}3 + 4 + 6 \\ 0 + 4 + 3 \\ 0 + 0 + 3\end{pmatrix} = \begin{pmatrix}13 \\ 7 \\ 3\end{pmatrix}.$$ Compare with $$\lambda \mathbf{v}_c = -2 \begin{pmatrix}1 \\ -2 \\ 3\end{pmatrix} = \begin{pmatrix}-2 \\ 4 \\ -6\end{pmatrix}$$. Since $$A\mathbf{v}_c \neq \lambda \mathbf{v}_c$$, $$\mathbf{v}_c$$ is not an eigenvector. - For $$\mathbf{v}_d$$: $$A\mathbf{v}_d = \begin{pmatrix}3 & -2 & 2 \\ 0 & -2 & 1 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}2 \\ 5 \\ 0\end{pmatrix} = \begin{pmatrix}(3)(2) + (-2)(5) + 2(0) \\ 0(2) + (-2)(5) + 1(0) \\ 0(2) + 0(5) + 1(0)\end{pmatrix} = \begin{pmatrix}6 - 10 + 0 \\ 0 - 10 + 0 \\ 0 + 0 + 0\end{pmatrix} = \begin{pmatrix}-4 \\ -10 \\ 0\end{pmatrix}.$$ Multiply $$\lambda \mathbf{v}_d = -2 \begin{pmatrix}2 \\ 5 \\ 0\end{pmatrix} = \begin{pmatrix}-4 \\ -10 \\ 0\end{pmatrix}$$. Since $$A\mathbf{v}_d = \lambda \mathbf{v}_d$$, $$\mathbf{v}_d$$ is an eigenvector corresponding to $$\lambda = -2$$. 4. **Final answer:** The eigenvector corresponding to the eigenvalue $$-2$$ is $$\boxed{\mathbf{v}_d = \begin{pmatrix}2 \\ 5 \\ 0\end{pmatrix}}.$$