Answer
$12$
Work Step by Step
We use one or a combination of theorems (11) to (15).
---
Starting with
$D=\left|\begin{array}{lll}{x}&{y}&{z}\\{u}&{v}&{w}\\{1}&{2}&{3}\end{array}\right|=4$,
Multiplying row $3$ with $3$, and adding it to row $1$ doesn't change the determinant (th.15).
So,
$D_{1}=\left|\begin{array}{lll}{x+3}&{y+6}&{z+9}\\{u}&{v}&{w}\\{1}&{2}&{3}\end{array}\right|=4$
Multiplying row $2$ with $3$ changes the determinant $D_{1}\rightarrow 3D_{1}$
(th 14)
$D_{2}=\left|\begin{array}{lll}{x+3}&{y+6}&{z+9}\\{3u}&{3v}&{3w}\\{1}&{2}&{3}\end{array}\right|=3(4)=12$
Multiplying row $3$ with $-1$, and adding it to row $2$ doesn't change the determinant (th.15).
So,
$D_{3}=\left|\begin{array}{lll}{x+3}&{y+6}&{z+9}\\{3u-1}&{3v-2}&{3w-3}\\{1}&{2}&{3}\end{array}\right|=12$
