1. Key Concepts and Formulas

• Coplanarity of Vectors: Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if and only if their scalar triple product is zero. Mathematically, $[\vec{a} \ \vec{b} \ \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.
• Scalar Triple Product as a Determinant: If $\vec{a} = a_1 \widehat i + a_2 \widehat j + a_3 \widehat k$, $\vec{b} = b_1 \widehat i + b_2 \widehat j + b_3 \widehat k$, and $\vec{c} = c_1 \widehat i + c_2 \widehat j + c_3 \widehat k$, their scalar triple product is given by the determinant: $[\vec{a} \ \vec{b} \ \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$
• Solving Polynomial Equations: The roots of a polynomial equation are the values of the variable that satisfy the equation. For a cubic equation, finding one root can help in factoring the polynomial into a linear and a quadratic term.

2. Step-by-Step Solution

Step 1: Formulate the condition for coplanarity. We are given three vectors: $\vec{v_1} = \mu \widehat i + \widehat j + \widehat k$ $\vec{v_2} = \widehat i + \mu \widehat j + \widehat k$ $\vec{v_3} = \widehat i + \widehat j + \mu \widehat k$

For these vectors to be coplanar, their scalar triple product must be zero. This can be represented by the determinant of the matrix formed by their components: $\begin{vmatrix} \mu & 1 & 1 \\ 1 & \mu & 1 \\ 1 & 1 & \mu \end{vmatrix} = 0$

Step 2: Expand the determinant. We expand the $3 \times 3$ determinant. Expanding along the first row: $\mu \left| \begin{matrix} \mu & 1 \\ 1 & \mu \end{matrix} \right| - 1 \left| \begin{matrix} 1 & 1 \\ 1 & \mu \end{matrix} \right| + 1 \left| \begin{matrix} 1 & \mu \\ 1 & 1 \end{matrix} \right| = 0$ Evaluate the $2 \times 2$ determinants: $\mu (\mu \cdot \mu - 1 \cdot 1) - 1 (1 \cdot \mu - 1 \cdot 1) + 1 (1 \cdot 1 - \mu \cdot 1) = 0$ $\mu (\mu^2 - 1) - (\mu - 1) + (1 - \mu) = 0$

Step 3: Simplify the equation to a polynomial in $\mu$. Distribute and combine terms: $\mu^3 - \mu - \mu + 1 + 1 - \mu = 0$ $\mu^3 - 3\mu + 2 = 0$

Step 4: Solve the cubic equation for $\mu$. We need to find the roots of the cubic equation $\mu^3 - 3\mu + 2 = 0$. We can test integer divisors of the constant term (2), which are $\pm 1, \pm 2$. Let's test $\mu = 1$: $(1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0$ Since $\mu = 1$ is a root, $(\mu - 1)$ is a factor of the polynomial. We can factor the polynomial by grouping or synthetic division. Let's use grouping: $\mu^3 - \mu - 2\mu + 2 = 0$ $\mu(\mu^2 - 1) - 2(\mu - 1) = 0$ $\mu(\mu - 1)(\mu + 1) - 2(\mu - 1) = 0$ Factor out the common term $(\mu - 1)$: $(\mu - 1)[\mu(\mu + 1) - 2] = 0$ $(\mu - 1)[\mu^2 + \mu - 2] = 0$ Now, we factor the quadratic term $\mu^2 + \mu - 2$: $\mu^2 + \mu - 2 = (\mu + 2)(\mu - 1)$ So the equation becomes: $(\mu - 1)(\mu + 2)(\mu - 1) = 0$ $(\mu - 1)^2 (\mu + 2) = 0$ The roots are $\mu = 1$ (with multiplicity 2) and $\mu = -2$.

Step 5: Identify the distinct real values of $\mu$ and calculate their sum. The values of $\mu$ for which the vectors are coplanar are $1, 1, -2$. The question asks for the sum of the distinct real values of $\mu$. The distinct real values are $1$ and $-2$. The sum of these distinct values is $1 + (-2) = -1$.

3. Common Mistakes & Tips

• Distinct Roots: Carefully read the question to ensure you are summing only the distinct real values of $\mu$, not all roots including multiplicities.
• Determinant Calculation: Double-check the signs and calculations during determinant expansion to avoid algebraic errors.
• Factoring Polynomials: If a root of a polynomial is found, use it to factor the polynomial. This simplifies finding all roots. For determinants of this specific symmetric structure, recognize the pattern for a faster solution.

4. Summary

The condition for three vectors to be coplanar is that their scalar triple product is zero. This condition translates into a determinant equation. Solving the resulting cubic equation $\mu^3 - 3\mu + 2 = 0$ yielded the roots $\mu = 1$ (with multiplicity 2) and $\mu = -2$. The distinct real values of $\mu$ are $1$ and $-2$. Their sum is $1 + (-2) = -1$.

The final answer is $\boxed{-1}$.