If the vectors $a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}, \hat{i}+\hat{j}+c \hat{k}$ $(a \neq b,…
If the vectors $a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}, \hat{i}+\hat{j}+c \hat{k}$ $(a \neq b, c \neq 1)$ are coplanar, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ has the value $\qquad$
1
-1
-2
5
Solution
Since $\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0$
Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1$ and $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1$, we get
$\begin{aligned}
& \left|\begin{array}{ccc}
a & 1 & 1 \\
1-a & b-1 & 0 \\
1-a & 0 & c-1
\end{array}\right|=0 \\
& \Rightarrow a(b-1)(c-1)-(1-a)(c-1)-(1-a)(b-1)=0
\end{aligned}$ Dividing by $(1-a)(1-b)(1-c)$,
we get $\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$
Consider, $\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}$
$=\frac{1}{1-a}-\frac{a}{1-a}$
....[From (i)]
$=1$