If `vecc=3veca-2vecb`then prove that`[(bara, barb, barc)]=0`

Explanation

To prove that \([ \vec{a}, \vec{b}, \vec{c} ] = 0\) given that \(\vec{c} = 3\vec{a} - 2\vec{b}\), we will follow these steps:

### Step 1: Understand the notation
The notation \([ \vec{a}, \vec{b}, \vec{c} ]\) represents the scalar triple product, which can be expressed as \(\vec{a} \cdot (\vec{b} \times \vec{c})\).

### Step 2: Substitute \(\vec{c}\)
We know that \(\vec{c} = 3\vec{a} - 2\vec{b}\). Therefore, we can write:
\[ [ \vec{a}, \vec{b}, \vec{c} ] = \vec{a} \cdot (\vec{b} \times (3\vec{a} - 2\vec{b})) \]

### Step 3: Expand the cross product
Using the distributive property of the cross product, we can expand this as:
\[ \vec{b} \times (3\vec{a} - 2\vec{b}) = \vec{b} \times 3\vec{a} - \vec{b} \times 2\vec{b} \]
This simplifies to:
\[ 3(\vec{b} \times \vec{a}) - 2(\vec{b} \times \vec{b}) \]

### Step 4: Simplify the terms
Since the cross product of any vector with itself is zero, we have:
\[ \vec{b} \times \vec{b} = 0 \]
Thus, the expression simplifies to:
\[ 3(\vec{b} \times \vec{a}) - 0 = 3(\vec{b} \times \vec{a}) \]

### Step 5: Substitute back into the scalar triple product
Now substituting this back into the scalar triple product:
\[ [ \vec{a}, \vec{b}, \vec{c} ] = \vec{a} \cdot (3(\vec{b} \times \vec{a})) \]
This can be factored as:
\[ = 3(\vec{a} \cdot (\vec{b} \times \vec{a})) \]

### Step 6: Evaluate the dot product
The dot product of a vector with the cross product of itself and another vector is always zero:
\[ \vec{a} \cdot (\vec{b} \times \vec{a}) = 0 \]
Thus, we have:
\[ [ \vec{a}, \vec{b}, \vec{c} ] = 3 \cdot 0 = 0 \]

### Conclusion
Therefore, we have shown that \([ \vec{a}, \vec{b}, \vec{c} ] = 0\).

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