If `vecu, vecv, vecw` are non -coplanar vectors and `p,q,` are real numbers then the equality
`[3vecu p vecv p vecw]-[p vecv vecw qvecu]-[2vecw-qvecv qvecu]=0` holds for

Options

A. exactly two values of `(p, q)`
B. more than two but not all values of `(p, q)`
C. all values of (p,q)
D. exactly one value of `(p, q)`

Answer

To solve the problem, we start with the given equation involving the vectors \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\):

\[ [3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w} - q\vec{v}, q\vec{u}] = 0 \]

### Step 1: Understand the notation
The notation \([ \cdot, \cdot, \cdot ]\) represents the scalar triple product of the vectors. The scalar triple product \([ \vec{a}, \vec{b}, \vec{c} ]\) is defined as \(\vec{a} \cdot (\vec{b} \times \vec{c})\).

### Step 2: Rewrite the equation
We can rewrite the equation using the properties of the scalar triple product. The scalar triple product is linear in each of its arguments. Thus, we can express the equation as:

\[ 3[\vec{u}, \vec{v}, \vec{w}] + p[\vec{v}, \vec{w}, \vec{u}] + p[\vec{w}, \vec{u}, \vec{v}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w} - q\vec{v}, q\vec{u}] = 0 \]

### Step 3: Expand the terms
Using the linearity of the scalar triple product, we can expand the terms:

1. \(3[\vec{u}, \vec{v}, \vec{w}]\)
2. \(p[\vec{v}, \vec{w}, \vec{u}]\) can be rewritten as \(p[\vec{u}, \vec{v}, \vec{w}]\) (by cyclic permutation).
3. \(p[\vec{w}, \vec{u}, \vec{v}]\) can also be rewritten as \(p[\vec{u}, \vec{v}, \vec{w}]\).
4. The term \([-p[\vec{v}, \vec{w}, \vec{u}]]\) can be rewritten as \(-p[\vec{u}, \vec{v}, \vec{w}]\).
5. The last term can be expanded as \([-2[\vec{w}, \vec{u}, \vec{v}] + q[\vec{v}, \vec{u}, \vec{w}]]\).

### Step 4: Combine like terms
Now we can combine all the terms:

\[ (3 + p - p - 2 + q)[\vec{u}, \vec{v}, \vec{w}] = 0 \]

This simplifies to:

\[ (1 + q)[\vec{u}, \vec{v}, \vec{w}] = 0 \]

### Step 5: Analyze the result
Since \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\) are non-coplanar vectors, the scalar triple product \([\vec{u}, \vec{v}, \vec{w}]\) cannot be zero. Therefore, we must have:

\[ 1 + q = 0 \implies q = -1 \]

### Step 6: Determine values for \(p\)
Now substituting \(q = -1\) back into the equation, we can analyze the quadratic equation formed. The equation simplifies to:

\[ 3p^2 - p(-1) + 2(-1)^2 = 0 \implies 3p^2 + p + 2 = 0 \]

### Step 7: Solve the quadratic equation
To find the values of \(p\), we can use the quadratic formula:

\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} \]

Calculating the discriminant:

\[ 1 - 24 = -23 \]

Since the discriminant is negative, there are no real solutions for \(p\).

### Conclusion
Thus, the equality holds for no values of \(p\) and \(q\).