For an $n\times n$ matrix $A$, a scalar $\lambda$ is an eigenvalue of $A$ if and only if it is a zero of the characteristic polynomial of $A$.
Why is this? Remember that $\lambda$ is an eigenvalue of $A$ if and only if there is a nonzero vector $\mathbf{v}$ such that $A\mathbf{v}=\lambda\mathbf{v}$; this is equivalent to the existence of a nonzero vector $\mathbf{v}$ such that $(A-\lambda I)\mathbf{v}=\mathbf{0}$. That means that the nullspace of the matrix $A-\lambda I$ (remember, $I$ is the identity matrix) is not just the zero vector, which means, necessarily, that $A-\lambda I$ is not invertible. Since it is not invertible, that means that its determinant is $0$; its determinant happens to equal the characteristic polynomial evaluated at $\lambda$, so this shows that if $\lambda$ is an eigenvalue of $A$, then $\lambda$ is a zero of the characteristic polynomial.
Conversely, if $\lambda$ is a zero of the characteristic polynomial of $A$, then the determinant of $A-\lambda I$ is zero, which means that $A-\lambda I$ is not invertible, which means there is a nonzero vector $\mathbf{w}$ such that $(A-\lambda I)\mathbf{w}=\mathbf{0}$. This shows that $\mathbf{w}$ is an eigenvector of $A$ with eigenvalue $\lambda$, so $\lambda$ is an eigenvalue.
For the matrix you have,
$$A = \left(\begin{array}{rrr}
3 & -1 & -1\\
-1 & 3 & -1\\
-1 & -1 & 3
\end{array}\right).$$
The characteristic polynomial $p(t)$ is:
$$\begin{align*}
p(t)=\det(A-tI) &= \left|\begin{array}{ccc}
3-t & -1 & -1\\
-1 & 3-t & -1\\
-1 & -1 & 3-t
\end{array}\right|\\
&= (3-t)\left|\begin{array}{cc}
3-t & -1\\
-1 & 3-t
\end{array}\right| +\left|\begin{array}{cc}
-1 & -1\\
-1 & 3-t
\end{array}\right| - \left|\begin{array}{cc}
-1 & -1\\
3-t & -1
\end{array}\right|\\
&= (3-t)\Bigl((3-t)^2-1\Bigr) + (t-4) - (4-t)\\
&= (3-t)\Bigl(t^2 -6t +8\Bigr) +2(t-4)\\
&= (3-t)(t-4)(t-2) + 2(t-4)\\
&= (t-4)\Bigl(2 - (t-2)(t-3)\Bigr) \\
&= -(t-4)(t^2-5t+6-2)\\
&= -(t-4)(t^2-5t+4)\\
&= -(t-4)^2(t-1).
\end{align*}$$
Since $\lambda$ is an eigenvalue of $A$ if and only if $p(\lambda)=0$, this says that the $A$ matrix has two distinct eigenvalues: $\lambda=4$, with algebraic multiplicity $2$, and $\lambda=1$.
What are the corresponding eigen vectors?
For $\lambda=1$, you want vectors $(a,b,c)^t$ such that $A(a,b,c)^t = (a,b,c)^t$ ($t$ is the transpose). Equivalently, you want the nullspace of $A-I$, except for $\mathbf{0}$. It is not hard to verify that $(1,1,1)^t$ is an eigenvector corresponding to $\lambda=1$, and that every eigenvector corresponding to $\lambda=1$ is a nonzero scalar multiple of $(1,1,1)^t$ (is this where you got confused? This is a vector, not a list of eigenvalues).
For $\lambda=4$, you want vectors $(a,b,c)^t$ that le in the nullspace of $A-4I$. Here, you want $a+b+c=0$, so the nullspace is spanned by the vectors $(1,0,-1)^t$ and $(0,1,-1)^t$; you can verify that each of these is an eigenvector corresponding to $\lambda=4$ and they are linearly independent, so the eigenvectors corresponding to $\lambda=4$ are the nonzero linear combinations of these two.
Best Answer
If the eigenvalues are distinct, then the characteristic polynomial is the only monic $n$th degree polynomial the matrix satisfies, since the eigenvalues are the roots, and so determine an $n$th degree polynomial up to a constant factor. If the minimal polynomial is the characteristic polynomial, we can make a similar statement. If not, then it's not true as ancientmathematician's answer shows.