Answer
$4$
Work Step by Step
We use one or a combination of theorems (11) to (15).
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Starting with
$D=\left|\begin{array}{lll}{x}&{y}&{z}\\{u}&{v}&{w}\\{1}&{2}&{3}\end{array}\right|$,
Interchanging rows 1 and 3 , we obtain
$D_{1}=\left|\begin{array}{lll} {1}&{2}&{3}\\{u}&{v}&{w}\\{x}&{y}&{z} \end{array}\right|=-D=-4,\quad$ by theorem (11).
Interchanging rows 2 and 3 of $D_{1}$, we obtain
$D_{2}=\left|\begin{array}{lll} {1}&{2}&{3}\\{x}&{y}&{z} \\{u}&{v}&{w} \end{array}\right|$
By theorem (11), $D_{2}=-D_{1}=-(-D)=D=4$
Adding row 3 to row 2 in $D_{2} $, we obtain
$D_{3}=\left|\begin{array}{ccc}{1}&{2}&{3}\\{x-u}&{y-v}&{z-w}\\{u}&{v}&{w}\end{array}\right|.$
By theorem (15), the determinant remains unchanged.
$D_{3}=4$
