Given Vectors: The given vectors are: \[ \vec{A} = a\hat{i} + \hat{j} + \hat{k} \] \[ \vec{B} = \hat{i} + b\hat{j} + \hat{k} \] \[ \vec{C} = \hat{i} + \hat{j} + c\hat{k} \] Since these vectors are coplanar, their scalar triple product must be zero, i.e., \[ \vec{A} \cdot (\vec{B} \times \vec{C}) = 0 \] Finding \(\vec{B} \times \vec{C}\): The cross-product determinant is: \[ \vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} \] Expanding along the first row: \[ \vec{B} \times \vec{C} = \hat{i} \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} \] \[ = \hat{i} (bc - 1) - \hat{j} (c - 1) + \hat{k} (1 - b) \] Computing \(\vec{A} \cdot (\vec{B} \times \vec{C})\): \[ \vec{A} \cdot (\vec{B} \times \vec{C}) = (a,1,1) \cdot (bc-1, -(c-1), 1-b) \] \[ = a(bc - 1) + 1(-c + 1) + 1(1 - b) \] \[ = abc - a - c + 1 + 1 - b \] \[ = abc - a - b - c + 2 = 0 \] \[ \Rightarrow abc - a - b - c = -2 \] Expressing in Terms of Given Sum: Dividing by \((1-a)(1-b)(1-c)\), we get: \[ \frac{abc - a - b - c}{(1-a)(1-b)(1-c)} = \frac{-2}{(1-a)(1-b)(1-c)} \] Rewriting: \[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \] Final Answer: \[ \boxed{1} \]