If `veca=2hati+3hatj+hatk, vecb=hati-2hatj+hatk` and `vecc=-3hati+hatj+2hatk`, then `[veca vecb vecc]=`

Explanation

To solve the problem, we need to calculate the scalar triple product of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). The scalar triple product can be calculated using the formula:

\[ [\vec{a} \, \vec{b} \, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) \]

### Step 1: Define the vectors
Given:
\[ \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} - 2\hat{j} + \hat{k} \]
\[ \vec{c} = -3\hat{i} + \hat{j} + 2\hat{k} \]

### Step 2: Calculate \(\vec{b} \times \vec{c}\)
To find \(\vec{b} \times \vec{c}\), we can use the determinant of a matrix formed by the unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) and the components of \(\vec{b}\) and \(\vec{c}\):

\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ -3 & 1 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -2 & 1 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ -3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ -3 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \(\begin{vmatrix} -2 & 1 \\ 1 & 2 \end{vmatrix} = (-2)(2) - (1)(1) = -4 - 1 = -5\)
2. \(\begin{vmatrix} 1 & 1 \\ -3 & 2 \end{vmatrix} = (1)(2) - (1)(-3) = 2 + 3 = 5\)
3. \(\begin{vmatrix} 1 & -2 \\ -3 & 1 \end{vmatrix} = (1)(1) - (-2)(-3) = 1 - 6 = -5\)

Putting it all together:
\[ \vec{b} \times \vec{c} = -5\hat{i} - 5\hat{j} - 5\hat{k} = -5(\hat{i} + \hat{j} + \hat{k}) \]

### Step 3: Calculate \(\vec{a} \cdot (\vec{b} \times \vec{c})\)
Now we calculate \(\vec{a} \cdot (\vec{b} \times \vec{c})\):
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = (2\hat{i} + 3\hat{j} + \hat{k}) \cdot (-5\hat{i} - 5\hat{j} - 5\hat{k}) \]

Calculating the dot product:
\[ = 2(-5) + 3(-5) + 1(-5) = -10 - 15 - 5 = -30 \]

### Conclusion
Thus, the scalar triple product \([\vec{a} \, \vec{b} \, \vec{c}] = -30\).