The sum of the distinct real values of μ for which the vectors, `muhat(i)+hat(j)+hat(k),hat(i)+muhat(j)+hat(k),hat(i)+hat(j)+muhat(k)` are co-planar is :

Options

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

Explanation

To determine 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 use the condition for coplanarity of three vectors. The vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are coplanar if the scalar triple product \( \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \).

### 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: Compute \( \vec{b} \times \vec{c} \)
To find \( \vec{b} \times \vec{c} \), we set up the determinant:

\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \mu & 1 \\ 1 & 1 & \mu \end{vmatrix} \]

Calculating the determinant, we get:

\[ \vec{b} \times \vec{c} = \hat{i} \left( \mu \cdot \mu - 1 \cdot 1 \right) - \hat{j} \left( 1 \cdot \mu - 1 \cdot 1 \right) + \hat{k} \left( 1 \cdot 1 - 1 \cdot \mu \right) \]

This simplifies to:

\[ \vec{b} \times \vec{c} = (\mu^2 - 1) \hat{i} - (\mu - 1) \hat{j} + (1 - \mu) \hat{k} \]

### Step 3: Compute \( \vec{a} \cdot (\vec{b} \times \vec{c}) \)
Now we compute the dot product \( \vec{a} \cdot (\vec{b} \times \vec{c}) \):

\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = (\mu \hat{i} + \hat{j} + \hat{k}) \cdot \left( (\mu^2 - 1) \hat{i} - (\mu - 1) \hat{j} + (1 - \mu) \hat{k} \right) \]

Calculating this, we have:

\[ = \mu(\mu^2 - 1) + 1(-(\mu - 1)) + 1(1 - \mu) \]

This simplifies to:

\[ = \mu^3 - \mu + 1 - \mu + 1 - \mu = \mu^3 - 3\mu + 2 \]

### Step 4: Set the equation to zero
For the vectors to be coplanar, we set the equation to zero:

\[ \mu^3 - 3\mu + 2 = 0 \]

### Step 5: Factor the cubic equation
To find the roots, we can use the Rational Root Theorem or synthetic division. Testing \( \mu = 1 \):

\[ 1^3 - 3(1) + 2 = 0 \]

So, \( \mu = 1 \) is a root. We can factor the cubic as:

\[ (\mu - 1)(\mu^2 + \mu - 2) = 0 \]

Now, we can factor \( \mu^2 + \mu - 2 \):

\[ \mu^2 + \mu - 2 = (\mu - 1)(\mu + 2) = 0 \]

Thus, the roots are \( \mu = 1 \) and \( \mu = -2 \).

### Step 6: Calculate the sum of distinct real 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 \( \mu \) for which the vectors are coplanar is \( \boxed{-1} \).