If your outer integral is over $z$ (you get to pick-the answer should come out the same) you need to figure out the range of $z$. In this case it runs from $0$ to $3$ so we have $\int_0^3 dz ($something). For the next integral, we get to consider $z$ to be fixed. If we do it over $y$, we need to figure out the range of $y$ at a given $z$. Now we have $\int_0^3 dz \int_{y_{min}}^{y_{max}} dy ($range of $x$ at this $(y,z))$. Now, considering $y$ and $z$ are fixed, you need to figure out the range of $x$, put that in the parentheses and you have your double integral.
Just to convert my comment to an answer.
If you want to use triple integral to find the volume. There are two ways to do this.
First method: Direct triple integral. You have to find the plane equations for all faces of this tetrahedron. Due to the piecewise nature, the limits of the integral are somewhat messy.
Second method: We can transform the points $ A(1,2,3)$, $B(-2,1,5)$, $C(3,7,1)$ to $A'(1,0,0)$, $B'(0,1,0)$, $C'(0,0,1)$. So that the volume of the new tetrahedron is easy to compute. The transformation matrix from $A',B',C'$ to $A,B,C$ is:
$$
T = \begin{pmatrix} 1&-2 &3
\\
2 &1 &7
\\
3 &5 &1\end{pmatrix} .
$$
Because the linearity this is also the Jacobian matrix, so
$$
\mathrm{Volume} = \iiint_{OABC} 1 \,dxdydz = \iiint_{OA'B'C'} 1 \,{|\det (T)|}\,dx'dy'dz' = \frac{17}{2}.
$$
Another tip with a formula to compute $n$-simplex, I took it from my computational geometry notes:
$$
|V| = \frac{1}{3!}\left|\det
\begin{pmatrix}
x_1 & x_2 & x_3 & x_4\\
y_1 & y_2 & y_3 & y_4\\
z_1 & z_2 & z_3 & z_4\\
1 & 1 & 1 & 1\\
\end{pmatrix}
\right|,
$$
where $(x_i,y_i,z_i)$ are the coordinates for the $i$-th vertex. It bears the same form for the area formula of a triangle with three vertices give
$$
|T| = \frac{1}{2!}\left|\det
\begin{pmatrix}
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3 \\
1 & 1 & 1 \\
\end{pmatrix}
\right|.
$$
Best Answer
HINT: If you are going to integrate with respect to $x$ and $y$, then the vectors $(0,2,0)-(0,0,1)=(0,2,-1)$ and $(2,2,0)-(0,0,1)=(2,2,-1)$ are in the plane you want. So $(0,2,-1)\times (2,2,-1)$ is a normal vector for the plane you want.