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

Options

A. 0.3
B. -0.3
C. 0.15
D. -0.15

Explanation

To find the scalar triple product \([ \vec{a} \, \vec{b} \, \vec{c} ]\), we will compute the determinant of the matrix formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

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 1: Write the vectors in matrix form
We will create a matrix with the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) from each vector:
\[ \begin{vmatrix} 2 & 3 & 1 \\ 1 & -2 & 1 \\ -3 & 1 & 2 \end{vmatrix} \]

### Step 2: Calculate the determinant
We will calculate the determinant using the formula for a 3x3 matrix:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

For our matrix:
- \(a = 2\), \(b = 3\), \(c = 1\)
- \(d = 1\), \(e = -2\), \(f = 1\)
- \(g = -3\), \(h = 1\), \(i = 2\)

Now, substituting these values into the determinant formula:
\[ \text{Det} = 2((-2)(2) - (1)(1)) - 3((1)(2) - (1)(-3)) + 1((1)(1) - (-2)(-3)) \]

Calculating each term:
1. First term:
\[ 2((-4) - 1) = 2(-5) = -10 \]
2. Second term:
\[ -3(2 + 3) = -3(5) = -15 \]
3. Third term:
\[ 1(1 - 6) = 1(-5) = -5 \]

Now, summing these results:
\[ \text{Det} = -10 - 15 - 5 = -30 \]

### Step 3: Final result
The scalar triple product \([ \vec{a} \, \vec{b} \, \vec{c} ]\) is:
\[ [ \vec{a} \, \vec{b} \, \vec{c} ] = -30 \]

### Summary
Thus, the final answer is:
\[ [ \vec{a} \, \vec{b} \, \vec{c} ] = -30 \]