$$\int\limits_0^1\int\limits_0^{1-x}\Bigg[\int\limits_{z=0}^{z=1-x-y}(1+x+y+z)^{-3}dz\Bigg]dxdy$$ $ \int\limits_0^1\int\limits_0^{1-x}\dfrac 1{-2}\Bigg[\dfrac 1{(1+x+y+z)^3}\Bigg]_0^{1-x-y}dxdy\\ \dfrac 1{-2} \int\limits_0^1\int\limits_0^{1-x}\Bigg[\dfrac 1{(1+x+y+1-x-y)^2}-\dfrac 1{(1+x+y)^2}\Bigg]dydx \\ \dfrac 1{-2} \int\limits_0^1\int\limits_0^{1-x}\Bigg[\dfrac 1{2^2}-\dfrac 1{(1+x+y)^2}\Bigg]dydx$

Now integrating w.r.t y we get

$$\dfrac 1{-2}\int\limits_0^1\Bigg[\dfrac y4-\dfrac 1{(1+x+y)^1}\times \dfrac 1{-1}\Bigg]_0^{1-x}dx$$

$ \dfrac 1{-2}\int\limits_0^1\Bigg[\dfrac{1-x}4 +\dfrac 1{1+x+1-x}-0-\dfrac 1{1+x}\Bigg]dx $

Now integrating w.r.t x

$$\dfrac 1{-2}\Bigg[\dfrac x4-\dfrac {x^2}8+\dfrac x2-\log(1+x)\Bigg]^1_0$$ Putting upper and lower limits We get $$=\dfrac 1{-2}\Bigg[\dfrac14-\dfrac 18+\dfrac 12-\log 2\Bigg]$$

$ =\dfrac 1{-2}\Bigg[\dfrac58-\log 2\Bigg]\\ =\dfrac 12\Bigg[\log 2-\dfrac 58\Bigg]$