If `vec(a)=2hat(i)+3hat(j)+hat(k), vec(b)=hat(i)-2hat(j)+hat(k)` and `vec(c )=-3hat(i)+hat(j)+2hat(k)`, find `[vec(a)vec(b)vec(c )]`.

Answer

To find the scalar triple product \([ \vec{a} \, \vec{b} \, \vec{c} ]\), we can use the determinant of a matrix formed by the components of 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 component form
The components of the vectors are:
- \(\vec{a} = (2, 3, 1)\)
- \(\vec{b} = (1, -2, 1)\)
- \(\vec{c} = (-3, 1, 2)\)

### Step 2: Form the matrix
We form a matrix using the components of the vectors:
\[ \begin{vmatrix} 2 & 3 & 1 \\ 1 & -2 & 1 \\ -3 & 1 & 2 \end{vmatrix} \]

### Step 3: Calculate the determinant
To find the determinant, we can use the formula for a \(3 \times 3\) matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

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

Calculating the determinant:
\[ = 2((-2)(2) - (1)(1)) - 3((1)(2) - (1)(-3)) + 1((1)(1) - (-2)(-3)) \]
\[ = 2(-4 - 1) - 3(2 + 3) + 1(1 - 6) \]
\[ = 2(-5) - 3(5) + 1(-5) \]
\[ = -10 - 15 - 5 \]
\[ = -30 \]

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

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