If xyz=(1-x)(1-y)(1-z) where 0 is less than equals to x,y,z is less than equals to 1 then the minimum value of x(1-z)+y(1-x)+z(1-y) is :


$$\begin{gathered} 2. If xyz = \left(1-x\right) \left(1-y\right) \left(1-z\right) where 0\le x,y,z\le 1, then the minimum value of x\left(1-z\right) +y\left(1-x\right) +z\left(1-y\right) is: \\ \left(a\right) \frac{3}{2} \left(b\right) \frac{1}{4} \left(c\right) \frac{3}{4} \left(d\right) \frac{1}{2} \end{gathered}$$