If `veca, vecb and vecc` are three vectors, such that `|veca|=2, |vecb|=3, |vecc|=4, veca. vecc=0, veca. vecb=0` and the angle between `vecb and vecc` is `(pi)/(3)`, then the value of `|veca xx (2vecb - 3 vecc)|` is equal to

Options

A. `12sqrt3`
B. `6sqrt3`
C. `3sqrt3`
D. 5

Answer

To solve the problem, we need to find the value of \(|\vec{a} \times (2\vec{b} - 3\vec{c})|\) given the conditions about the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

### Step-by-step Solution:

1. **Understanding the Given Information**:
- \(|\vec{a}| = 2\)
- \(|\vec{b}| = 3\)
- \(|\vec{c}| = 4\)
- \(\vec{a} \cdot \vec{c} = 0\) (implying \(\vec{a}\) is perpendicular to \(\vec{c}\))
- \(\vec{a} \cdot \vec{b} = 0\) (implying \(\vec{a}\) is perpendicular to \(\vec{b}\))
- The angle between \(\vec{b}\) and \(\vec{c}\) is \(\frac{\pi}{3}\).

2. **Finding the Magnitude of the Cross Product**:
We need to compute \(|\vec{a} \times (2\vec{b} - 3\vec{c})|\). Since \(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\), it is also perpendicular to any linear combination of these vectors, including \(2\vec{b} - 3\vec{c}\).

Therefore, we can use the property of the cross product:
\[ |\vec{a} \times (2\vec{b} - 3\vec{c})| = |\vec{a}| \cdot |2\vec{b} - 3\vec{c}| \]

3. **Calculating \(|2\vec{b} - 3\vec{c}|\)**:
To find \(|2\vec{b} - 3\vec{c}|\), we use the formula for the magnitude of the difference of two vectors:
\[ |2\vec{b} - 3\vec{c}| = \sqrt{|2\vec{b}|^2 + |-3\vec{c}|^2 - 2(2\vec{b}) \cdot (-3\vec{c})} \]

- Calculate \(|2\vec{b}| = 2|\vec{b}| = 2 \times 3 = 6\)
- Calculate \(|-3\vec{c}| = 3|\vec{c}| = 3 \times 4 = 12\)
- Now, calculate the dot product \(2\vec{b} \cdot (-3\vec{c})\):
\[ 2\vec{b} \cdot (-3\vec{c}) = -6(\vec{b} \cdot \vec{c}) = -6|\vec{b}||\vec{c}|\cos\left(\frac{\pi}{3}\right) \]
Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\):
\[ \vec{b} \cdot \vec{c} = 3 \cdot 4 \cdot \frac{1}{2} = 6 \]
Therefore,
\[ 2\vec{b} \cdot (-3\vec{c}) = -6 \times 6 = -36 \]

4. **Putting it all together**:
Now we can substitute back into the magnitude formula:
\[ |2\vec{b} - 3\vec{c}| = \sqrt{6^2 + 12^2 - (-36)} \]
\[ = \sqrt{36 + 144 + 36} = \sqrt{216} = 6\sqrt{6} \]

5. **Final Calculation**:
Now, substituting back into the equation for the cross product:
\[ |\vec{a} \times (2\vec{b} - 3\vec{c})| = |\vec{a}| \cdot |2\vec{b} - 3\vec{c}| = 2 \cdot 6\sqrt{6} = 12\sqrt{6} \]

### Final Answer:
\[ |\vec{a} \times (2\vec{b} - 3\vec{c})| = 12\sqrt{6} \]