In matrix from, the system is
[tex]\dfrac{\mathrm d}{\mathrm dt}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3&-1&-1\\1&1&-1\\1&-1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}[/tex]
The coefficient matrix has eigenvalues [tex]\lambda[/tex] such that
[tex]\begin{vmatrix}3-\lambda&-1&-1\\1&1-\lambda&-1\\1&-1&1-\lambda\end{vmatrix}=-(\lambda-2)^2(\lambda-1)=0\implies\lambda=2_{(2)},\lambda=1[/tex]
(where the subscript denotes multiplicity of the eigenvalue).
[tex]\lambda=2[/tex]:
[tex]\begin{bmatrix}1&-1&-1\\1&-1&-1\\1&-1&-1\end{bmatrix}\vec\eta_1=\vec0[/tex]
[tex]\implies\eta_{1,1}-\eta_{1,2}-\eta_{1,3}=0\implies\eta_{1,1}=\eta_{1,2}+\eta_{1,3}[/tex]
By picking [tex]\eta_{1,1}=1[/tex], we can then set [tex]\eta_{1,2}=1[/tex] and [tex]\eta_{1,3}=0[/tex], and vice versa, to find two corresponding eigenvectors,
[tex]\vec\eta_1=\begin{bmatrix}1\\1\\0\end{bmatrix},\vec\eta_2=\begin{bmatrix}1\\0\\1\end{bmatrix}[/tex]
[tex]\lambda=1[/tex]:
[tex]\begin{bmatrix}2&-1&-1\\1&0&-1\\1&-1&0\end{bmatrix}\vec\eta_3=\vec0[/tex]
[tex]\implies2\eta_{3,1}-\eta_{3,2}-\eta_{3,3}=0\implies2\eta_{3,1}=\eta_{3,2}+\eta_{3,3}[/tex]
We obtain [tex]\vec\eta_3[/tex] independent of [tex]\vec\eta_1,\vec\eta_2[/tex] by picking [tex]\eta_{3,2}=\eta_{3,3}=1[/tex], so that the third corresponding eigenvector is
[tex]\vec\eta_3=\begin{bmatrix}1\\1\\1\end{bmatrix}[/tex]
Then the general solution to this system is
[tex]\begin{bmatrix}x\\y\\z\end{bmatrix}=C_1e^{2t}\vec\eta_1+C_2te^{2t}\vec\eta_2+C_3e^t\vec\eta_3[/tex]
[tex]\begin{cases}x=C_1e^{2t}+C_2te^{2t}+C_3e^t\\y=C_1e^{2t}+C_3e^t\\z=C_2te^{2t}+C_3e^t\end{cases}[/tex]
