Let `veca=hati-hatj+hatk, vecb=2hati+hatj+hatk and vecc=hati+hatj-2hatk`, then the value of `[(veca, vecb, vecc)]` is equal to

Answer

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

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

### Step 2: Write the coefficients of the vectors
The coefficients of the vectors are:
- For \(\vec{a}\): \(1, -1, 1\)
- For \(\vec{b}\): \(2, 1, 1\)
- For \(\vec{c}\): \(1, 1, -2\)

### Step 3: Set up the determinant
We can form a determinant using these coefficients:
\[ \text{Det} = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & -2 \end{vmatrix} \]

### Step 4: Calculate the determinant
Using the formula for the determinant of a 3x3 matrix:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]
we can substitute:
- \(a = 1, b = -1, c = 1\)
- \(d = 2, e = 1, f = 1\)
- \(g = 1, h = 1, i = -2\)

Calculating the determinant:
\[ = 1 \cdot (1 \cdot (-2) - 1 \cdot 1) - (-1) \cdot (2 \cdot (-2) - 1 \cdot 1) + 1 \cdot (2 \cdot 1 - 1 \cdot 1) \]

Calculating each term:
1. \(1 \cdot (-2 - 1) = 1 \cdot (-3) = -3\)
2. \(-(-1) \cdot (-4 - 1) = 1 \cdot (-5) = -5\)
3. \(1 \cdot (2 - 1) = 1 \cdot 1 = 1\)

Putting it all together:
\[ \text{Det} = -3 - 5 + 1 = -7 \]

### Final Result
Thus, the value of the scalar triple product \([( \vec{a}, \vec{b}, \vec{c} )]\) is:
\[ \boxed{-7} \]