Key Concepts and Formulas

• Coplanarity of Vectors: Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if and only if their scalar triple product is zero, i.e., $[\vec{u} \ \vec{v} \ \vec{w}] = 0$.
• Scalar Triple Product using Determinants: If vectors are given in component form $\vec{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}$, $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$, and $\vec{w} = w_1\hat{i} + w_2\hat{j} + w_3\hat{k}$, then their scalar triple product is given by the determinant of their components: $\begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}$
• Algebraic Manipulation of Fractions: To add fractions, find a common denominator and combine the numerators.

Step-by-Step Solution

Step 1: Apply the condition for coplanarity. The given vectors are $\vec{u} = a \hat{i}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$, $\vec{v} = \hat{\mathrm{i}}+b \hat{j}+\hat{\mathrm{k}}$, and $\vec{w} = \hat{\mathrm{i}}+\hat{\mathrm{j}}+c \hat{\mathrm{k}}$. For these vectors to be coplanar, their scalar triple product must be zero. We express this condition using a determinant of their components: $\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$

Step 2: Expand the determinant to obtain an algebraic relation. We expand the determinant along the first row: $a \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + 1 \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} = 0$ Calculate the $2 \times 2$ determinants: $a(bc - 1 \cdot 1) - 1(1 \cdot c - 1 \cdot 1) + 1(1 \cdot 1 - b \cdot 1) = 0$ $a(bc - 1) - (c - 1) + (1 - b) = 0$ Distribute and simplify: $abc - a - c + 1 + 1 - b = 0$ $abc - a - b - c + 2 = 0$ This equation is the fundamental relationship between $a, b, c$ that arises from the coplanarity of the given vectors.

Step 3: Manipulate the algebraic relation to simplify the target expression. We need to find the value of $S = \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$. To combine these fractions, we find a common denominator, which is $(1-a)(1-b)(1-c)$: $S = \frac{(1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b)}{(1-a)(1-b)(1-c)}$ Let's expand the numerator ($N$) and the denominator ($D$) separately.

Numerator ($N$): $N = (1 - b - c + bc) + (1 - a - c + ac) + (1 - a - b + ab)$ $N = 1 - b - c + bc + 1 - a - c + ac + 1 - a - b + ab$ Combine like terms: $N = (1+1+1) - (a+a) - (b+b) - (c+c) + (ab+bc+ca)$ $N = 3 - 2a - 2b - 2c + ab + bc + ca$ $N = 3 - 2(a+b+c) + (ab+bc+ca)$

Denominator ($D$): $D = (1-a)(1-b)(1-c)$ Expand $(1-a)(1-b)$: $(1-a)(1-b) = 1 - b - a + ab$ Now multiply by $(1-c)$: $D = (1 - a - b + ab)(1 - c)$ $D = 1(1-c) - a(1-c) - b(1-c) + ab(1-c)$ $D = 1 - c - a + ac - b + bc + ab - abc$ Group like terms: $D = 1 - (a+b+c) + (ab+bc+ca) - abc$

Step 4: Substitute the coplanarity relation into the denominator. From Step 2, we have the relation $abc - a - b - c + 2 = 0$. We can rearrange this to express $abc$ in terms of $a, b, c$: $abc = a+b+c-2$ Substitute this expression for $abc$ into the expanded denominator $D$: $D = 1 - (a+b+c) + (ab+bc+ca) - (a+b+c-2)$ $D = 1 - a - b - c + ab + bc + ca - a - b - c + 2$ Combine like terms: $D = (1+2) - (a+a) - (b+b) - (c+c) + (ab+bc+ca)$ $D = 3 - 2a - 2b - 2c + ab + bc + ca$ $D = 3 - 2(a+b+c) + (ab+bc+ca)$

Step 5: Evaluate the target expression. Now we have the numerator $N = 3 - 2(a+b+c) + (ab+bc+ca)$ and the denominator $D = 3 - 2(a+b+c) + (ab+bc+ca)$. We can see that $N = D$. The expression we need to evaluate is $S = \frac{N}{D}$. Since $a, b, c$ are distinct real numbers and none are equal to one, the denominators $(1-a), (1-b), (1-c)$ are non-zero, which means $D \neq 0$. Therefore, $S = \frac{D}{D} = 1$

Common Mistakes & Tips

• Determinant Calculation Errors: Be meticulous with the signs when expanding the determinant. A single sign error can lead to an incorrect algebraic relation.
• Algebraic Simplification: The process of expanding and simplifying the numerator and denominator of the target expression requires careful attention to detail. Grouping terms systematically helps avoid errors.
• Using the Coplanarity Condition: Ensure the coplanarity condition is correctly translated into a determinant and then used to simplify the final expression. The condition $a, b, c \neq 1$ is crucial for the expression to be well-defined.

Summary

The problem requires applying the condition of coplanarity for three vectors, which translates to a zero scalar triple product and thus a zero determinant of their components. Expanding this determinant yields an algebraic relationship between $a, b, c$. The core of the problem then lies in algebraic manipulation: combining the fractions in the target expression and using the derived relationship to simplify the numerator and denominator. It turns out that the numerator and denominator are equal, leading to a value of 1 for the expression.

The final answer is \boxed{1}.