Answer
4
Work Step by Step
We know:
i) Switching any two rows or columns causes the determinant to switch sign.
$\implies$ doing this twice causes the determinant to stay the same.
ii) Adding one row to another row does not change the determinant.
$\begin{vmatrix}
1&2&3\\
x-u&y-v&z-w\\
u&v&w
\end{vmatrix}$
Performing $R_{2} \leftrightarrow R_{1}$, then $R_{3} \leftrightarrow R_{1}$, by rule (i), the determinant stays the same.
$\implies\begin{vmatrix}
1&2&3\\
x-u&y-v&z-w\\
u&v&w
\end{vmatrix} \longrightarrow \begin{vmatrix}
x-u&y-v&z-w\\
u&v&w\\
1&2&3
\end{vmatrix}$
Performing $R_{1}+R_{2}$, by rule (ii), the determinant stays the same.
$\implies \begin{vmatrix}
x-u&y-v&z-w\\
u&v&w\\
1&2&3
\end{vmatrix} \longrightarrow \begin{vmatrix}
x&y&z\\
u&v&w\\
1&2&3
\end{vmatrix}$
Given that,
$\begin{vmatrix}
x&y&z\\
u&v&w\\
1&2&3
\end{vmatrix} = 4$
$\therefore Answer = 4$
