Coplanarity Using Determinant

Since the vectors are coplanar, their scalar triple product is zero:

$$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$$

Apply Row Operations to isolate $$(1-a), (1-b), (1-c)$$:

Perform $$R_2 \to R_2 - R_1$$ and $$R_3 \to R_3 - R_1$$:

$$\begin{vmatrix} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \end{vmatrix} = 0$$

To make it look like our target expression, let's factor out $$(1-a)$$ from $$R_2$$ and $$R_3$$ (or directly divide elements):

$$a[(b-1)(c-1) - 0] - 1[(1-a)(c-1) - 0] + 1[0 - (1-a)(b-1)] = 0$$

$$a(1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b) = 0$$

Divide the entire equation by $$(1-a)(1-b)(1-c)$$:

$$\frac{a(1-b)(1-c)}{(1-a)(1-b)(1-c)} + \frac{(1-a)(1-c)}{(1-a)(1-b)(1-c)} + \frac{(1-a)(1-b)}{(1-a)(1-b)(1-c)} = 0$$

$$\frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$$

Notice that $$\frac{a}{1-a} = \frac{a - 1 + 1}{1-a} = \frac{-(1-a) + 1}{1-a} = -1 + \frac{1}{1-a}$$.

$$-1 + \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$$

$$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$$