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
$$
|T| = \frac{1}{2!}\left|\det
\begin{pmatrix}
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3 \\
1 & 1 & 1 \\
\end{pmatrix}
\right|.
$$
To get an idea what to do in 3D, try to understand the 2D case first and try to generalize. When integrating over the triangle with vertices $(0,0)$, $(0,1)$ and $(1,0)$, it is often a good idea to first let $x$ go from zero to $1-y$ and then let $y$ go from zero to 1.
In your case, you can proceed analogously: let $x$ range in $[0,1-y-z]$, then $y$ in $[0,1-z]$ and finally $z$ in $[0,1]$.
That is,
$$
\int_{\text{tetrahedron}}f(x,y,z)dxdydz
=
\int_0^1\int_0^{1-z}\int_0^{1-y-z}f(x,y,z)dxdydz.
$$
In your case $f(x,y,z)=xyz$.
To make the iterated integral structure clear, you can write it as
$$
\int_0^1\left(\int_0^{1-z}\left(\int_0^{1-y-z}xyzdx\right)dy\right)dz
$$
and start by doing the innermost integral.
The first integral is $\int_0^{1-y-z}xyzdx=\frac12yz(1-y-z)^2$.
If you get this right, you are on the right track.
Best Answer
Your limits are correct, considering the YZ axis, if you want another example to understand better I have given the limits considering the XZ plane.
The tetrahedron is having its base on the XZ plane and YZ plane so it will be a good idea to consider the limits from the XZ plane to the $x+y=z$ plane.
considering the limits from the YZ plane to the $x+y=z$
$y=0$ to $y=z-x$
Now as the limits for the volume has been taken care of, the area formed on the XZ plane is considered. If the strip is considered parallel to the z axis then the limits
$z=x$ to $z=1$ and the limits of the x are $x=0$ to $x=1$ . This limit has been chosen to make it easier to solve the last integral using the Beta function.
The equation will be $$\int_0^1\int_x^{1}\int_0^{z-x}dy dz dx$$