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.