The sum of the distinct real values of `mu`, for which the vectors, `mu hati + hatj + hatk, hati + mu hatj + hatk, hati + hatj + muhatk` are coplanar, is

Options

A. 2
B. 0
C. 1
D. `-1`

Answer

To solve the problem of finding the sum of the distinct real values of \( \mu \) for which the vectors \( \mu \hat{i} + \hat{j} + \hat{k} \), \( \hat{i} + \mu \hat{j} + \hat{k} \), and \( \hat{i} + \hat{j} + \mu \hat{k} \) are coplanar, we can follow these steps:

### Step 1: Define the vectors
Let:
- \( \vec{A} = \mu \hat{i} + \hat{j} + \hat{k} \)
- \( \vec{B} = \hat{i} + \mu \hat{j} + \hat{k} \)
- \( \vec{C} = \hat{i} + \hat{j} + \mu \hat{k} \)

### Step 2: Use the condition for coplanarity
The vectors \( \vec{A}, \vec{B}, \vec{C} \) are coplanar if their scalar triple product is zero:
\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = 0 \]

### Step 3: Set up the determinant
The scalar triple product can be computed using the determinant of a matrix formed by the coefficients of the vectors:
\[ \begin{vmatrix} \mu & 1 & 1 \\ 1 & \mu & 1 \\ 1 & 1 & \mu \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Expanding the determinant:
\[ = \mu \begin{vmatrix} \mu & 1 \\ 1 & \mu \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \mu \end{vmatrix} + 1 \begin{vmatrix} 1 & \mu \\ 1 & 1 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ = \mu (\mu^2 - 1) - (1 \cdot \mu - 1 \cdot 1) + (1 \cdot 1 - \mu \cdot 1) \]
\[ = \mu (\mu^2 - 1) - (\mu - 1) + (1 - \mu) \]
\[ = \mu^3 - \mu - \mu + 1 + 1 - \mu \]
\[ = \mu^3 - 3\mu + 2 \]

### Step 5: Set the determinant to zero
Setting the expression to zero gives us:
\[ \mu^3 - 3\mu + 2 = 0 \]

### Step 6: Factor the polynomial
To find the roots of the polynomial, we can use the Rational Root Theorem or synthetic division. Testing \( \mu = 1 \):
\[ 1^3 - 3(1) + 2 = 0 \]
So \( \mu - 1 \) is a factor. Dividing \( \mu^3 - 3\mu + 2 \) by \( \mu - 1 \):
\[ \mu^3 - 3\mu + 2 = (\mu - 1)(\mu^2 + \mu - 2) \]

### Step 7: Factor the quadratic
Now, we need to factor \( \mu^2 + \mu - 2 \):
\[ \mu^2 + 2\mu - \mu - 2 = (\mu + 2)(\mu - 1) \]
Thus, we have:
\[ (\mu - 1)^2(\mu + 2) = 0 \]

### Step 8: Find the distinct real values of \( \mu \)
The roots are:
- \( \mu = 1 \) (with multiplicity 2)
- \( \mu = -2 \)

### Step 9: Calculate the sum of distinct values
The distinct real values of \( \mu \) are \( 1 \) and \( -2 \). Therefore, the sum is:
\[ 1 + (-2) = -1 \]

### Final Answer
The sum of the distinct real values of \( \mu \) is \( \boxed{-1} \).