If `veca+vecb ,vecc ` are any three non- coplanar vectors then the equation `[vecbxxvecc veccxxveca vecaxxvecb]x^(2) + [veca+vecbvecb+veccvecc+veca] x+1 +[vecb-veccvecc -vecc-vecaveca -vecb] =0` has roots

Options

A. real and distinct
B. real
C. equal
D. imaginary

Explanation

To solve the given problem, we need to analyze the equation involving the vectors and determine the nature of its roots. Let's break down the solution step by step.

### Step 1: Understand the Given Equation
The equation is given as:
\[ [\vec{b} \times \vec{c}, \vec{c} \times \vec{a}, \vec{a} \times \vec{b}] x^2 + [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] x + 1 + [\vec{b} - \vec{c}, \vec{c} - \vec{a}, \vec{a} - \vec{b}] = 0 \]
We need to simplify this expression.

### Step 2: Simplify the First Term
The first term can be expressed using the scalar triple product:
\[ [\vec{b} \times \vec{c}, \vec{c} \times \vec{a}, \vec{a} \times \vec{b}] = [\vec{b}, \vec{c}, \vec{a}]^2 \]
Let \( A = [\vec{a}, \vec{b}, \vec{c}] \). Then:
\[ [\vec{b} \times \vec{c}, \vec{c} \times \vec{a}, \vec{a} \times \vec{b}] = A^2 \]

### Step 3: Simplify the Second Term
The second term can also be simplified:
\[ [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] = [\vec{a}, \vec{b}, \vec{c}] + [\vec{b}, \vec{c}, \vec{a}] = 2A \]

### Step 4: Simplify the Third Term
The third term can be simplified as follows:
\[ [\vec{b} - \vec{c}, \vec{c} - \vec{a}, \vec{a} - \vec{b}] = [\vec{b}, \vec{c}, \vec{a}] - [\vec{c}, \vec{a}, \vec{b}] = 0 \]

### Step 5: Substitute Back into the Equation
Now substituting these simplifications back into the equation gives:
\[ A^2 x^2 + 2A x + 1 = 0 \]

### Step 6: Find the Discriminant
To determine the nature of the roots, we calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (2A)^2 - 4(A^2)(1) = 4A^2 - 4A^2 = 0 \]

### Step 7: Conclusion
Since the discriminant \( D = 0 \), the quadratic equation has equal roots.

### Final Answer
Therefore, the equation has equal roots.

---