Let $I$ be the $3\times 3$ identity matrix, $u=(1,1,1)^T$, $v=(a,b,c)^T$, and $w=(x,y,z)^T$.
Then we wish to evaluate the determinant of $I+v u^T + u w^T$.
It is as easy to do this for dimension $n$ as dimension $3$ so let's do that. The shape of this matrix (and indeed the shape of the proposed answer) suggest that we should look at a larger matrix, one where the identity is 'bordered' in some way by $u,v,w$. After a little experimentation it seems we should look at the $(n+1+1)\times (n+1+1)$ block matrix
$$M=
\begin{bmatrix}
I & u & v\\
w^T &-1 &0\\
u^T & 0 & -1\\
\end{bmatrix}.
$$
We can add the $(n+1)$st row to each of the first $n$ rows without changing the value of the determinant, and likewise add the $(n+2)$nd column to each of the first $n$ columns without changing the value of the determinant. That is,
$$
\det\begin{bmatrix}
I & u & v\\
w^T &-1 &0\\
u^T & 0 & -1\\
\end{bmatrix}
=
\det\begin{bmatrix}
I+u w^T+v u^T & 0 & v\\
w^T &-1 &0\\
0 & 0 & -1\\
\end{bmatrix}
=
\det(I+u w^T+v u^T).
$$
However we can evaluate $\det M$ in a different way. We can subtract the sum of the first $n$ columns from the $(n+1)$st column without changing the value of the determinant, and likewise we can subtract the sum of the first $n$ columns, each weighted by the corresponding entry of $v$ from the last column without changing the value of the determinant. That is,
$$
\det\begin{bmatrix}
I & u & v\\
w^T &-1 &0\\
u^T & 0 & -1\\
\end{bmatrix}
=
\det\begin{bmatrix}
I& 0 & 0\\
w^T &-1-u^T w &-v^T w\\
u^T & -u^T u & -1-v^T u\\
\end{bmatrix}
=
(1+u^T w)(1+w^T u)-n v^T w.
$$
In the $3\times 3$ case that gives
$$
\det
\begin{bmatrix}
1 +a +x & a+y & a+z\\
b +x & 1+ b+y & b+z\\
c +x & c+y & 1+c+z\\
\end{bmatrix}
=
(1+x+y+z)(1+a+b+c)-3(ax+by+cz).
$$
[I suspect this is just some variant of the Matrix Determinant Lemma.]
