Let `vec(a),vec(b)` and `vec(c )` be three non-coplaner vectors and `vec(p),vec(q),vec(r)` be three vectors such that `vec(p)=2vec(a)-vec(b)+vec(c ),vec(q)=vec(a)-3vec(b)+2vec(c),vec(r)=vec(a)+vec(b)-vec(c )`. If `[vec(a)vec(b)vec(c)]=2` then `[vec(p)vec(q)vec( r)]` equals

Options

A. 6
B. -6
C. 3
D. None of these

Explanation

To solve the problem, we need to find the scalar triple product \([ \vec{p}, \vec{q}, \vec{r} ]\) given the vectors \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\) expressed in terms of the non-coplanar vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). We know that \([ \vec{a}, \vec{b}, \vec{c} ] = 2\).

### Step 1: Write the expressions for \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\)

Given:
\[ \vec{p} = 2\vec{a} - \vec{b} + \vec{c} \]
\[ \vec{q} = \vec{a} - 3\vec{b} + 2\vec{c} \]
\[ \vec{r} = \vec{a} + \vec{b} - \vec{c} \]

### Step 2: Compute \(\vec{p} \times \vec{q}\)

To find \([ \vec{p}, \vec{q}, \vec{r} ]\), we first need to compute \(\vec{p} \times \vec{q}\):
\[ \vec{p} \times \vec{q} = (2\vec{a} - \vec{b} + \vec{c}) \times (\vec{a} - 3\vec{b} + 2\vec{c}) \]

Using the distributive property of the cross product:
\[ \vec{p} \times \vec{q} = 2\vec{a} \times \vec{a} - 6\vec{a} \times \vec{b} + 4\vec{a} \times \vec{c} - \vec{b} \times \vec{a} + 3\vec{b} \times \vec{b} - 2\vec{b} \times \vec{c} + \vec{c} \times \vec{a} - 3\vec{c} \times \vec{b} + 2\vec{c} \times \vec{c} \]

Since the cross product of any vector with itself is zero:
\[ \vec{p} \times \vec{q} = -6\vec{a} \times \vec{b} + 4\vec{a} \times \vec{c} - \vec{b} \times \vec{a} - 2\vec{b} \times \vec{c} + \vec{c} \times \vec{a} - 3\vec{c} \times \vec{b} \]

### Step 3: Simplify the expression

Using the property \(\vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})\):
\[ \vec{p} \times \vec{q} = -6\vec{a} \times \vec{b} + 4\vec{a} \times \vec{c} + \vec{a} \times \vec{b} - 2\vec{b} \times \vec{c} + \vec{c} \times \vec{a} - 3\vec{b} \times \vec{c} \]

Combining like terms:
\[ \vec{p} \times \vec{q} = (-6 + 1)\vec{a} \times \vec{b} + (4 + 1)\vec{a} \times \vec{c} + (-2 - 3)\vec{b} \times \vec{c} \]
\[ = -5\vec{a} \times \vec{b} + 5\vec{a} \times \vec{c} - 5\vec{b} \times \vec{c} \]

### Step 4: Compute \([ \vec{p}, \vec{q}, \vec{r} ]\)

Now we compute:
\[ [ \vec{p}, \vec{q}, \vec{r} ] = \vec{p} \times \vec{q} \cdot \vec{r} \]
Substituting \(\vec{r} = \vec{a} + \vec{b} - \vec{c}\):
\[ [ \vec{p}, \vec{q}, \vec{r} ] = (-5\vec{a} \times \vec{b} + 5\vec{a} \times \vec{c} - 5\vec{b} \times \vec{c}) \cdot (\vec{a} + \vec{b} - \vec{c}) \]

Calculating the dot products:
1. \((-5\vec{a} \times \vec{b}) \cdot \vec{a} = 0\)
2. \((-5\vec{a} \times \vec{b}) \cdot \vec{b} = 0\)
3. \((-5\vec{a} \times \vec{b}) \cdot (-\vec{c}) = 5(\vec{a} \times \vec{b}) \cdot \vec{c} = 5[ \vec{a}, \vec{b}, \vec{c} ] = 10\)

4. \((5\vec{a} \times \vec{c}) \cdot \vec{a} = 0\)
5. \((5\vec{a} \times \vec{c}) \cdot \vec{b} = 5(\vec{a} \times \vec{c}) \cdot \vec{b} = -5[ \vec{a}, \vec{b}, \vec{c} ] = -10\)
6. \((5\vec{a} \times \vec{c}) \cdot (-\vec{c}) = 0\)

7. \((-5\vec{b} \times \vec{c}) \cdot \vec{a} = -5(\vec{b} \times \vec{c}) \cdot \vec{a} = 5[ \vec{a}, \vec{b}, \vec{c} ] = 10\)
8. \((-5\vec{b} \times \vec{c}) \cdot \vec{b} = 0\)
9. \((-5\vec{b} \times \vec{c}) \cdot (-\vec{c}) = 0\)

### Step 5: Combine results

Combining all the contributions:
\[ [ \vec{p}, \vec{q}, \vec{r} ] = 10 - 10 + 10 = 10 \]

### Final Result

Thus, the value of \([ \vec{p}, \vec{q}, \vec{r} ]\) is:
\[ \boxed{10} \]