If the vectors a hat{i}+hat{j}+hat{k},hat{i}+ | vedclass

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(B) Since the vectors are coplanar,their scalar triple product is zero: $\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vmatrix} = 0$ Applyi

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$1$

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(B) Since the vectors are coplanar,their scalar triple product is zero: $\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vmatrix} = 0$ Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$: $\begin{vmatrix} a & 1 & 1 \ 1-a & b-1 & 0 \ 1-a & 0 & c-1 \end{vmatrix} = 0$ Expanding along the first row: $a((b-1)(c-1) - 0) - 1((1-a)(c-1) - 0) + 1(0 - (1-a)(b-1)) = 0$ $a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$ Divide the entire equation by $(1-a)(1-b)(1-c)$ (noting $a, b, c 
eq 1$): $\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$ $\frac{-a}{(1-a)} - \frac{1}{(1-b)} - \frac{1}{(1-c)} = 0$ Since $\frac{-a}{1-a} = \frac{1-a-1}{1-a} = 1 - \frac{1}{1-a}$,we substitute: $1 - \frac{1}{1-a} - \frac{1}{1-b} - \frac{1}{1-c} = 0$ Therefore,$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$.

If the vectors $a \hat{i}+\hat{j}+\hat{k}$,$\hat{i}+b \hat{j}+\hat{k}$,and $\hat{i}+\hat{j}+c \hat{k}$ $(a \neq 1, b \neq 1, c \neq 1, a \neq b \neq c)$ are coplanar,then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is:

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