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

Answer

To find the scalar triple product \([ \vec{a} \vec{b} \vec{c} ]\), we will use the determinant of a 3x3 matrix formed by the components of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

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

### Step 1: Write the vectors in component form
The vectors can be represented as:
- \(\vec{a} = (1, -2, 3)\)
- \(\vec{b} = (2, -3, 1)\)
- \(\vec{c} = (3, 1, -2)\)

### Step 2: Set up the determinant
The scalar triple product can be expressed as the determinant of the following matrix:
\[ \begin{vmatrix} 1 & -2 & 3 \\ 2 & -3 & 1 \\ 3 & 1 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we can use the formula for a 3x3 determinant:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Where:
- \(a = 1\), \(b = -2\), \(c = 3\)
- \(d = 2\), \(e = -3\), \(f = 1\)
- \(g = 3\), \(h = 1\), \(i = -2\)

Calculating each part:
1. \(ei - fh = (-3)(-2) - (1)(1) = 6 - 1 = 5\)
2. \(di - fg = (2)(-2) - (1)(3) = -4 - 3 = -7\)
3. \(dh - eg = (2)(1) - (-3)(3) = 2 + 9 = 11\)

Now substituting back into the determinant formula:
\[ \text{Det} = 1(5) - (-2)(-7) + 3(11) \]
\[ = 5 - 14 + 33 \]
\[ = 5 - 14 + 33 = 24 \]

### Final Result:
Thus, the value of the scalar triple product \([ \vec{a} \vec{b} \vec{c} ]\) is \(24\).

---