If `veca, vecb` and `vecc` are any three vectors forming a linearly independent system, then `AA theta in R`
`vecp=vecacostheta+vecbsintheta+vecc(cos2theta)`
`vecq=vecacos((2pi)/(3)+theta)+vecbsin((2pi)/(3)+theta) + vecc(cos 2)((2pi)/(3)+theta)`
and `vecr=vecacos(theta-(2pi)/(3))+vecbsin(theta-(2pi)/(3))+vecc cos2(theta-(2pi)/(3))`
then `[vecpvecqvecr]`

Options

A. `[vecavecbvecc]costheta`
B. `[vecavecbvecc]cos2theta`
C. `[vecavecbvecc]cos3theta`
D. None of these

Explanation

To solve the problem, we need to analyze the vectors \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\) given in the question.

### Step 1: Express the vectors
We have the following vectors:
\[ \vec{p} = \vec{a} \cos \theta + \vec{b} \sin \theta + \vec{c} \cos 2\theta \]
\[ \vec{q} = \vec{a} \cos\left(\frac{2\pi}{3} + \theta\right) + \vec{b} \sin\left(\frac{2\pi}{3} + \theta\right) + \vec{c} \cos 2\left(\frac{2\pi}{3} + \theta\right) \]
\[ \vec{r} = \vec{a} \cos\left(\theta - \frac{2\pi}{3}\right) + \vec{b} \sin\left(\theta - \frac{2\pi}{3}\right) + \vec{c} \cos 2\left(\theta - \frac{2\pi}{3}\right) \]

### Step 2: Show that \(\vec{p} + \vec{q} + \vec{r} = 0\)
We need to show that the sum of these vectors is zero:
\[ \vec{p} + \vec{q} + \vec{r} = 0 \]

To do this, we can use the angle addition formulas for sine and cosine.

1. For \(\vec{q}\):
- \(\cos\left(\frac{2\pi}{3} + \theta\right) = -\frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta\)
- \(\sin\left(\frac{2\pi}{3} + \theta\right) = \frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta\)
- \(\cos 2\left(\frac{2\pi}{3} + \theta\right) = \cos\left(2\theta + \frac{4\pi}{3}\right) = -\cos(2\theta) - \frac{1}{2}\)

2. For \(\vec{r}\):
- \(\cos\left(\theta - \frac{2\pi}{3}\right) = -\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta\)
- \(\sin\left(\theta - \frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta\)
- \(\cos 2\left(\theta - \frac{2\pi}{3}\right) = \cos\left(2\theta - \frac{4\pi}{3}\right) = -\cos(2\theta) + \frac{1}{2}\)

After substituting these into \(\vec{p} + \vec{q} + \vec{r}\) and simplifying, we find that all terms cancel out, leading to:
\[ \vec{p} + \vec{q} + \vec{r} = 0 \]

### Step 3: Conclusion about coplanarity
Since \(\vec{p} + \vec{q} + \vec{r} = 0\), it implies that the vectors \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\) are coplanar.

### Step 4: Volume of the parallelepiped
The volume of the parallelepiped formed by three vectors is given by the scalar triple product \([\vec{p} \vec{q} \vec{r}]\). Since the vectors are coplanar, the volume is zero:
\[ [\vec{p} \vec{q} \vec{r}] = 0 \]

### Final Answer
Thus, the final answer is:
\[ [\vec{p} \vec{q} \vec{r}] = 0 \]