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