Let $a, b, c$ be three distinct real numbers, none equal to one. If the vectors $$a \hat{i}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+b \hat{j}+\hat{\mathrm{k}}$$ and $\hat{\mathrm{i}}+\hat{\mathrm{j}}+c \hat{\mathrm{k}}$ are coplanar, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to :

$$ \left|\begin{array}{lll} a & 1 & 1 \\\\ 1 & \mathrm{~b} & 1 \\\\ 1 & 1 & \mathrm{c} \end{array}\right|=0 $$ <br/><br/>$$ \mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1, \mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1 $$ <br/><br/>$$ \begin{aligned} & \left|\begin{array}{lll} a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1 \end{array}\right|=0 \\\\ & a(b-1)(c-1)-(1-a)(c-1)+(1-a)(1-b)=0 \\\\ & a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \frac{\mathrm{a}}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\\\ & \Rightarrow-1+\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\\\ & \Rightarrow \frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=1 \end{aligned} $$