Question

If $(a,1,1),(1,b,1),(1,1,c)$ are coplanar, prove that $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$

Answer 1

Given that $(a,1,1),(1,b,1),(1,1,c)$ are coplanar.

Therefore, $\left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0$

Now perform column operations, $C_1\to C_1-C_3;C_2\to C_2-C_3$

$\Rightarrow \left|\begin{array}{ccc}a-1& 0& 1\\ 0& b-1& 1\\ 1-c& 1-c& c\end{array}\right|=0$

$\Rightarrow (a-1)(b-1)(c-1)\left|\begin{array}{ccc}1& 0& \frac{1}{a-1}\\ 0& 1& \frac{1}{b-1}\\ 1& 1& \frac{c}{1-c}\end{array}\right|=0$

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

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

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

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

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