1. The problem is to evaluate the triple integral $$\int_0^1 \int_0^1 \int_0^1 \log x \log y \log z \, dz \, dx \, dy.$$ 2. The integral is over the unit cube $[0,1]^3$ and the integrand is the product of logarithms of each variable. Since the variables are independent and the limits are constants, the triple integral can be separated into the product of three single integrals: $$\int_0^1 \log x \, dx \times \int_0^1 \log y \, dy \times \int_0^1 \log z \, dz.$$ 3. Recall the formula for the integral of $\log t$ over $[0,1]$: $$\int_0^1 \log t \, dt = -1.$$ This can be derived by integration by parts: let $u=\log t$, $dv=dt$, then $du=\frac{1}{t} dt$, $v=t$, so $$\int_0^1 \log t \, dt = [t \log t]_0^1 - \int_0^1 t \cdot \frac{1}{t} dt = 0 - \int_0^1 1 dt = -1.$$ 4. Applying this to each integral, we get: $$\int_0^1 \log x \, dx = -1,$$ $$\int_0^1 \log y \, dy = -1,$$ $$\int_0^1 \log z \, dz = -1.$$ 5. Therefore, the value of the triple integral is: $$(-1) \times (-1) \times (-1) = -1.$$ 6. In conclusion, the triple integral evaluates to $$\boxed{-1}.$$