Solution:

$I=\int_{x=0}^{1} \int_{y=0}^{1-X} \int_{z=0}^{1-x-y}(x+y+z) d z\ d y\ d x$

$I=\int_{x=0}^{1} \int_{y=0}^{1-x}\left[\frac{(x+y+z)^{2}}{2}\right]_{0}^{1-x-y} d y\ d z$

$I=\frac{1}{2} \int_{x=0}^{1} \int_{y=0}^{1-x}\left[1-(x+y)^{2}\right] d y \ d x$

$\begin{aligned} I &=\frac{1}{2} \int_{x=0}^{1}\left[y-\frac{(x+y)^{2}}{2}\right]_{0}^{1-x} d x \\ I &=\frac{1}{2} \int_{x=0}^{1}\left[(1-x)-\frac{1}{3}+\frac{x^{3}}{3}\right] d x \\ I &=\frac{1}{2}\left[\frac{2 x}{3}-\frac{x^{2}}{2}+\frac{x^{4}}{12}\right]_{0}^{1} \\ I &=\frac{1}{2} \cdot \frac{3}{12}=\frac{1}{8} \\ \therefore I &=\frac{1}{8} \end{aligned}$