If `veca,vecb` and `vecc` are unit coplanar vectors, then
`[(2veca-3vecb,7vecb-9vecc,12vecc-23vecb)]` is equal to

Options

A. 0
B. `1//2`
C. `24`
D. `32`

Explanation

To solve the problem, we need to evaluate the scalar triple product of the vectors \( (2\vec{a} - 3\vec{b}, 7\vec{b} - 9\vec{c}, 12\vec{c} - 23\vec{b}) \).

### Step-by-step Solution:

1. **Identify the Vectors**:
Let:
\[ \vec{u} = 2\vec{a} - 3\vec{b} \]
\[ \vec{v} = 7\vec{b} - 9\vec{c} \]
\[ \vec{w} = 12\vec{c} - 23\vec{b} \]

2. **Set Up the Scalar Triple Product**:
We need to compute the scalar triple product \( [\vec{u}, \vec{v}, \vec{w}] \), which can be expressed as:
\[ [\vec{u}, \vec{v}, \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w}) \]

3. **Calculate the Cross Product \( \vec{v} \times \vec{w} \)**:
Using the distributive property of the cross product:
\[ \vec{v} \times \vec{w} = (7\vec{b} - 9\vec{c}) \times (12\vec{c} - 23\vec{b}) \]
Expanding this:
\[ = 7\vec{b} \times 12\vec{c} - 7\vec{b} \times 23\vec{b} - 9\vec{c} \times 12\vec{c} + 9\vec{c} \times 23\vec{b} \]
Since the cross product of any vector with itself is zero:
\[ \vec{b} \times \vec{b} = 0 \quad \text{and} \quad \vec{c} \times \vec{c} = 0 \]
Therefore, we simplify to:
\[ = 84(\vec{b} \times \vec{c}) + 207(\vec{c} \times \vec{b}) \]
Since \( \vec{c} \times \vec{b} = -(\vec{b} \times \vec{c}) \):
\[ = 84(\vec{b} \times \vec{c}) - 207(\vec{b} \times \vec{c}) = (84 - 207)(\vec{b} \times \vec{c}) = -123(\vec{b} \times \vec{c}) \]

4. **Calculate the Dot Product \( \vec{u} \cdot (\vec{v} \times \vec{w}) \)**:
Now we compute:
\[ \vec{u} \cdot (-123(\vec{b} \times \vec{c})) = -123(\vec{u} \cdot (\vec{b} \times \vec{c})) \]
Substitute \( \vec{u} = 2\vec{a} - 3\vec{b} \):
\[ = -123((2\vec{a} - 3\vec{b}) \cdot (\vec{b} \times \vec{c})) \]
Using the property of scalar triple products:
\[ \vec{u} \cdot (\vec{b} \times \vec{c}) = [\vec{a}, \vec{b}, \vec{c}] \]
Since \( \vec{a}, \vec{b}, \vec{c} \) are coplanar, we have:
\[ [\vec{a}, \vec{b}, \vec{c}] = 0 \]

5. **Final Result**:
Therefore, the scalar triple product evaluates to:
\[ -123 \cdot 0 = 0 \]

### Conclusion:
The value of the scalar triple product \( [(2\vec{a} - 3\vec{b}, 7\vec{b} - 9\vec{c}, 12\vec{c} - 23\vec{b})] \) is **0**.