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

Options

A. 30
B. -30
C. 15
D. -15

Explanation

To solve the problem, we need to find the scalar triple product of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). The scalar triple product can be calculated using the determinant of a matrix formed by 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 1: Set up the determinant
We will create a 3x3 matrix using the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) from the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\):

\[ \begin{vmatrix} 2 & 3 & 1 \\ 1 & -2 & 1 \\ -3 & 1 & 2 \end{vmatrix} \]

### Step 2: Calculate the determinant
We can calculate the determinant using the formula for a 3x3 matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

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:

\[ D = 2((-2)(2) - (1)(1)) - 3((1)(2) - (1)(-3)) + 1((1)(1) - (-2)(-3)) \]

Calculating each term:

1. First term:
\[ 2((-2)(2) - (1)(1)) = 2(-4 - 1) = 2(-5) = -10 \]

2. Second term:
\[ -3((1)(2) - (1)(-3)) = -3(2 + 3) = -3(5) = -15 \]

3. Third term:
\[ 1((1)(1) - (-2)(-3)) = 1(1 - 6) = 1(-5) = -5 \]

### Step 3: Combine the results
Now we combine the results from the three terms:

\[ D = -10 - 15 - 5 = -30 \]

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

### Final Answer:
\[ [\vec{a} \, \vec{b} \, \vec{c}] = -30 \]

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