If `vec(a)` is perpendicular to `vec(b)` and `vec( c ),| vec(a) |=2, |vec(b)|=3,|vec(c )|=4` and angle between `vec(b)` and `vec( c)` is `( 2pi )/( 3)`, then `[ vec(a)"" vec( b) "" vec( c)]` is equal to

Options

A. `4 sqrt(3)`
B. `6 sqrt(3)`
C. `12 sqrt(3)` or `- 12 sqrt(3)`
D. `18 sqrt(3)`

Answer

To solve the problem, we need to find the scalar triple product (STP) of vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). Given that \(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\), we can use the formula for the scalar triple product:

\[ [\vec{a}, \vec{b}, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) \]

### Step 1: Identify the magnitudes and angle
We know:
- \(|\vec{a}| = 2\)
- \(|\vec{b}| = 3\)
- \(|\vec{c}| = 4\)
- The angle between \(\vec{b}\) and \(\vec{c}\) is \(\frac{2\pi}{3}\).

### Step 2: Calculate \(\vec{b} \times \vec{c}\)
The magnitude of the cross product \(\vec{b} \times \vec{c}\) can be calculated using the formula:

\[ |\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin(\theta) \]

where \(\theta\) is the angle between \(\vec{b}\) and \(\vec{c}\). Substituting the values:

\[ |\vec{b} \times \vec{c}| = 3 \cdot 4 \cdot \sin\left(\frac{2\pi}{3}\right) \]

### Step 3: Calculate \(\sin\left(\frac{2\pi}{3}\right)\)
The sine of \(\frac{2\pi}{3}\) is:

\[ \sin\left(\frac{2\pi}{3}\right) = \sin\left(180^\circ - 60^\circ\right) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Step 4: Substitute back to find \(|\vec{b} \times \vec{c}|\)
Now substituting the value of \(\sin\left(\frac{2\pi}{3}\right)\):

\[ |\vec{b} \times \vec{c}| = 3 \cdot 4 \cdot \frac{\sqrt{3}}{2} = 6\sqrt{3} \]

### Step 5: Calculate the scalar triple product
Now we can find the scalar triple product:

\[ [\vec{a}, \vec{b}, \vec{c}] = |\vec{a}| \cdot |\vec{b} \times \vec{c}| \]

Substituting the values:

\[ [\vec{a}, \vec{b}, \vec{c}] = 2 \cdot 6\sqrt{3} = 12\sqrt{3} \]

### Final Answer
Thus, the scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\) is:

\[ \boxed{12\sqrt{3}} \]