The points $(0,0,0),$ $(a,1,1),$ $(1,b,1),$ and $(1,1,c)$ are coplanar, where $a \neq 1,$ $b \neq 1,$ and $c \neq 1.$  Compute
\[\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}.\]
The normal vector to the plane will be orthogonal to both
\[\begin{pmatrix} a \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ b \\ 1 \end{pmatrix} = \begin{pmatrix} a - 1 \\ 1 - b \\ 0 \end{pmatrix}\]and
\[\begin{pmatrix} a \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ c \end{pmatrix} = \begin{pmatrix} a - 1 \\ 0 \\ 1 - c \end{pmatrix}.\]Their cross product is
\[\begin{pmatrix} a - 1 \\ 1 - b \\ 0 \end{pmatrix} \times \begin{pmatrix} a - 1 \\ 0 \\ 1 - c \end{pmatrix} = \begin{pmatrix} (b - 1)(c - 1) \\ (a - 1)(c - 1) \\ (a - 1)(b - 1) \end{pmatrix}.\]By scaling, we can take $\begin{pmatrix} 1/(1 - a) \\ 1/(1 - b) \\ 1/(1 - c) \end{pmatrix}$ as the normal vector.  Since the plane passes through $(0,0,0),$ the equation of the plane is
\[\frac{x}{1 - a} + \frac{y}{1 - b} + \frac{z}{1 - c} = 0.\]Since the plane passes through $(a,1,1),$
\[\frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 0.\]Adding 1 to both sides, we get
\[\frac{a}{1 - a} + 1 + \frac{1}{1 - b} + \frac{1}{1 - c} = 1,\]so
\[\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = \boxed{1}.\]