If `veca,vecb, vecc` are unit coplanar vectors then the scalar triple product `[2veca-vecb, 2vecb-c ,vec2c-veca]` is equal to (A) `0` (B) `1` (C) `-sqrt(3)` (D) `sqrt(3)`

Options

A. 0
B. 1
C. `-sqrt3`
D. `sqrt3`

Explanation

To solve the problem, we need to evaluate the scalar triple product \([2\vec{a} - \vec{b}, 2\vec{b} - \vec{c}, 2\vec{c} - \vec{a}]\) given that \(\vec{a}, \vec{b}, \vec{c}\) are unit coplanar vectors.

### Step-by-step Solution:

1. **Understanding the Scalar Triple Product**:
The scalar triple product of three vectors \(\vec{u}, \vec{v}, \vec{w}\) is given by \(\vec{u} \cdot (\vec{v} \times \vec{w})\). It represents the volume of the parallelepiped formed by the three vectors. If the vectors are coplanar, the volume is zero.

2. **Identify the Vectors**:
Let:
\[ \vec{u} = 2\vec{a} - \vec{b}, \quad \vec{v} = 2\vec{b} - \vec{c}, \quad \vec{w} = 2\vec{c} - \vec{a} \]

3. **Calculate the Scalar Triple Product**:
We need to compute:
\[ [\vec{u}, \vec{v}, \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w}) \]

4. **Finding \(\vec{v} \times \vec{w}\)**:
First, calculate \(\vec{v} \times \vec{w}\):
\[ \vec{v} \times \vec{w} = (2\vec{b} - \vec{c}) \times (2\vec{c} - \vec{a}) \]
Using the distributive property of the cross product:
\[ = 2\vec{b} \times 2\vec{c} - 2\vec{b} \times \vec{a} - \vec{c} \times 2\vec{c} + \vec{c} \times \vec{a} \]
Since \(\vec{c} \times \vec{c} = \vec{0}\), we have:
\[ = 4\vec{b} \times \vec{c} - 2\vec{b} \times \vec{a} + \vec{c} \times \vec{a} \]

5. **Finding \(\vec{u} \cdot (\vec{v} \times \vec{w})\)**:
Now compute \(\vec{u} \cdot (\vec{v} \times \vec{w})\):
\[ \vec{u} \cdot (\vec{v} \times \vec{w}) = (2\vec{a} - \vec{b}) \cdot (4\vec{b} \times \vec{c} - 2\vec{b} \times \vec{a} + \vec{c} \times \vec{a}) \]
Expanding this using the distributive property:
\[ = 2\vec{a} \cdot (4\vec{b} \times \vec{c}) - \vec{b} \cdot (4\vec{b} \times \vec{c}) - 2\vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{a}) \]
The term \(\vec{b} \cdot (4\vec{b} \times \vec{c}) = 0\) since a vector dotted with a cross product of itself and another vector is zero. The term \(\vec{a} \cdot (\vec{b} \times \vec{a}) = 0\) for the same reason.

6. **Final Expression**:
Thus, we have:
\[ = 8\vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{c} \times \vec{a}) \]
Since \(\vec{a}, \vec{b}, \vec{c}\) are coplanar, \(\vec{a} \cdot (\vec{b} \times \vec{c}) = 0\). Hence:
\[ \vec{u} \cdot (\vec{v} \times \vec{w}) = 0 \]

### Conclusion:
The scalar triple product \([2\vec{a} - \vec{b}, 2\vec{b} - \vec{c}, 2\vec{c} - \vec{a}]\) is equal to \(0\).

### Answer:
(A) \(0\)