Answer
$0$
Work Step by Step
Let us consider a $ 3\times 3$ determinant with all the elements in the first column be $0$.
The first column of the array of signs is $ + , - , +$ .
$$\begin{vmatrix}
0 & a & d\\
0 & b & c\\
0 & e&f
\end{vmatrix}=0\begin{vmatrix}
b & c \\
e & f
\end{vmatrix}-0\begin{vmatrix}
a& d \\
e& f
\end{vmatrix}+0\begin{vmatrix}
a & d \\
b& c
\end{vmatrix}$$
$$=0(bf-ec)-0(af-de)+0(ac-db)=0-0+0=0$$
Note that the determinant of the matrix will be the same regardless of the column you select to expand by.
The result is the same if the determinant has $n$ rows and $n$ columns.