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

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(A) Since the vectors are coplanar,their scalar triple product is zero: $\left|\begin{array}{lll}a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c\end{array}ight|

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$1$

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(A) Since the vectors are coplanar,their scalar triple product is zero: $\left|\begin{array}{lll}a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c\end{array}ight|=0$. Applying $R_2 ightarrow R_2-R_1$ and $R_3 ightarrow R_3-R_1$,we get $\left|\begin{array}{ccc} a & 1 & 1 \ 1-a & b-1 & 0 \ 1-a & 0 & c-1 \end{array}ight|=0$. Expanding along the first row: $a(b-1)(c-1) - 1(1-a)(c-1) + 1(0 - (b-1)(1-a)) = 0$. $a(b-1)(c-1) + (1-a)(c-1) + (1-a)(b-1) = 0$. Dividing the entire equation by $(1-a)(1-b)(1-c)$,we get: $\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} + \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} + \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$. This simplifies to: $\frac{-a}{(1-a)} + \frac{1}{(1-b)} + \frac{1}{(1-c)} = 0$. We can write $\frac{-a}{1-a}$ as $\frac{1-a-1}{1-a} = 1 - \frac{1}{1-a}$. Substituting this back: $1 - \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$. Therefore,$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$.

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

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