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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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(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 column operations $C_2 ightarrow C_2 - C_1$ and $C_3 ightarrow C_3 - C_1$:$\left|\begin{array}{lll}a & 1-a & 1-a \ 1 & b-1 & 0 \ 1 & 0 & c-1\end{array}ight|=0$Expanding the determinant along the first row:$a(b-1)(c-1) - (1-a)(c-1) + (1-a)(1-b) = 0$Divide the entire equation by $(1-a)(1-b)(1-c)$ (noting that $a, b, c 
eq 1$):$\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)(1-b)}{(1-a)(1-b)(1-c)} = 0$$\frac{a}{(1-a)} + \frac{1}{(1-b)} + \frac{1}{(1-c)} = 0$Since $\frac{a}{1-a} = \frac{a-1+1}{1-a} = -1 + \frac{1}{1-a}$,we substitute this into the equation:$-1 + \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$Therefore,$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$.

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

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