= Solution
Put $q=(1,1,1,1)^{\mathsf T}$ and arrange the four vectors as the columns of $A=(x-1)I+qq^{\mathsf T}$. On the one-dimensional <vector subspace> spanned by $q$, the <eigenvalue> is $x+3$. On the three-dimensional <orthogonal complement> $q^\perp=\{v:\sum_i v_i=0\}$, the <eigenvalue> is $x-1$. Thus
$$
\det A=(x+3)(x-1)^3,
$$
and \b[the vectors fail to be a basis exactly when $x=1$ or $x=-3$.]
For $x=1$, all four columns equal $q$. The <matrix rank> is $1$, a <basis> of their <span> is $\{q\}$, and an extension to a <basis> of $\mathbb R^4$ is
$$
\boxed{\{q,e_1,e_2,e_3\}.}
$$
The last coordinate first forces the coefficient of $q$ to vanish in any <linear dependence>, after which the other coefficients vanish.
For $x=-3$, the <matrix rank> is $3$ and the <span> is exactly $q^\perp$. A <basis> consists of the first three original columns,
$$
\boxed{\{(-3,1,1,1),(1,-3,1,1),(1,1,-3,1)\}.}
$$
Indeed, writing them as $q-4e_i$, a vanishing linear combination has coefficient sum zero by its fourth coordinate, and then each coefficient is zero by the first three coordinates. Adjoin $q$ to extend this <basis> to $\mathbb R^4$, since $q\notin q^\perp$.
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