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
$$
|T| = \frac{1}{2!}\left|\det
\begin{pmatrix}
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3 \\
1 & 1 & 1 \\
\end{pmatrix}
\right|.
$$
The equation of the plane containing the points $(2,0,0),(0,2,0),(0,0,1)$ is given by $x+y+2z-2=0,$ that is, $z=1-\frac x2-\frac y2.$ Note that the basis of the tetrahedron is a triangle, one side on the $OX$-axis, one side on the $OY$-axis, and the equation of the third one is given by $x+y=2.$ Thus, the volume can be obtained by evaluating the integral
$$\int_0^2\int_0^{2-x} \left(1-\frac x2-\frac y2\right)dydx. $$
We have:
$$\int_0^2\int_0^{2-x} \left(1-\frac x2-\frac y2\right)dydx=\int_0^2 \left(\left(1-\frac x2\right)(2-x)-\frac{(2-x)^2}{4}\right)dx \\ =\int_0^2 \left(1-x+\frac{x^2}{4}\right)dx. $$
Best Answer
Well I would first make a sketch of the solid boundry. By getting three points from the relation x+y+z=1 by plugging in 0 for two of the variables and solving for the other. You get the points (1,0,0), (0,0,1),(0,1,0). I would sketch that first, then put the plane equation in terms of two variables, I.e. z=1-x-y. Thus you can see that z goes from 0 to 1-x-y. In addition the bounds for x and y are determined by its projection in the xy plane, a triangle, with the hypotenuse y=1-x.