Surface Integrals - Parametric Representation 
Find the parametric representation for the parts of the plane $$2x+3y+z=4$$ where $$1\leq x+y+z\leq 7$$ and $$2\leq x-y\leq4$$.

My attempt:
I thought to let $u=x+y+z$ and $v=x-y$ such that $1\leq u \leq 7$ and $2\leq v\leq4$. But I'm unable to find a suitable parametric representation.

All the points that lie on the relevant parts of the plane satisfy the following equations, 
\begin{align*}
2x + 3y + z &= 4 \\
x+y+z &= u \\
x-y &= v
\end{align*}
where $1 \leq u \leq 7$ and $2 \leq v \leq 4$.
If we can find expressions for $x$, $y$ and $z$ in terms of $u$ and $v$, we have found a parametrisation. Doing so involves solving the system of three linear equations above for $x$, $y$ and $z$. To do this, we can solve the equation $$\begin{bmatrix}
    2       & 3 & 1 \\
    1       & 1 & 1  \\
    1       & -1 & 0 
\end{bmatrix} \begin{bmatrix}
    x \\
    y\\
    z  
\end{bmatrix}=
\begin{bmatrix}
    4 \\
    u\\
    v  
\end{bmatrix}$$
The solution ends up being $$\begin{bmatrix}
    x \\
    y\\
    z  
\end{bmatrix} = \begin{bmatrix}
    \frac{-u+2v+4}{3} \\
    \frac{-u-v+4}{3}\\
    \frac{5u-v-8}{3}  
\end{bmatrix}$$
Hence, we can say that your parametrisation is defined by the function $\textbf{x}:\mathbb{R}^2 \rightarrow \mathbb{R}^3$ given by $$\textbf{x}(u,v) = \begin{bmatrix}
    \frac{-u+2v+4}{3} \\
    \frac{-u-v+4}{3}\\
    \frac{5u-v-8}{3}  
\end{bmatrix}$$ Where $1 \leq u \leq 7$ and $2 \leq v \leq 4$.
