To solve the problem, we need to find the maximum value of the expression \([ \vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a} ]\), where \(\vec{a} = \hat{i} + \hat{j} + 2\hat{k}\), \(\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}\), and \(|\vec{c}| = 1\).
### Step 1: Calculate \(\vec{a} \times \vec{b}\)
We will first compute the cross product \(\vec{a} \times \vec{b}\):
\[
\vec{a} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}
\]
Using the determinant method for the cross product:
\[
\vec{a} \times \vec{b} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 2 \\
1 & 2 & 2
\end{vmatrix}
\]
Calculating the determinant:
\[
= \hat{i} \begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix}
\]
Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix} = (1)(2) - (2)(2) = 2 - 4 = -2\)
2. \(\begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = (1)(2) - (2)(1) = 2 - 2 = 0\)
3. \(\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = (1)(2) - (1)(1) = 2 - 1 = 1\)
Putting it all together:
\[
\vec{a} \times \vec{b} = -2\hat{i} - 0\hat{j} + 1\hat{k} = -2\hat{i} + \hat{k}
\]
### Step 2: Calculate \(\vec{b} \times \vec{c}\)
Next, we need to compute \(\vec{b} \times \vec{c}\). Since \(|\vec{c}| = 1\), we can express \(\vec{c}\) in terms of its components as \(\vec{c} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\) where \(x^2 + y^2 + z^2 = 1\).
Using the same determinant method:
\[
\vec{b} \times \vec{c} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 2 \\
x & y & z
\end{vmatrix}
\]
Calculating this determinant yields:
\[
= \hat{i} \begin{vmatrix} 2 & 2 \\ y & z \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ x & z \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ x & y \end{vmatrix}
\]
Calculating the 2x2 determinants:
1. \(\begin{vmatrix} 2 & 2 \\ y & z \end{vmatrix} = 2z - 2y = 2(z - y)\)
2. \(\begin{vmatrix} 1 & 2 \\ x & z \end{vmatrix} = z - 2x\)
3. \(\begin{vmatrix} 1 & 2 \\ x & y \end{vmatrix} = y - 2x\)
Thus:
\[
\vec{b} \times \vec{c} = 2(z - y)\hat{i} - (z - 2x)\hat{j} + (y - 2x)\hat{k}
\]
### Step 3: Calculate \(\vec{c} \times \vec{a}\)
Now we compute \(\vec{c} \times \vec{a}\):
\[
\vec{c} \times \vec{a} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
x & y & z \\
1 & 1 & 2
\end{vmatrix}
\]
Calculating this determinant yields:
\[
= \hat{i} \begin{vmatrix} y & z \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} x & z \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} x & y \\ 1 & 1 \end{vmatrix}
\]
Calculating the 2x2 determinants:
1. \(\begin{vmatrix} y & z \\ 1 & 2 \end{vmatrix} = 2y - z\)
2. \(\begin{vmatrix} x & z \\ 1 & 2 \end{vmatrix} = 2x - z\)
3. \(\begin{vmatrix} x & y \\ 1 & 1 \end{vmatrix} = y - x\)
Thus:
\[
\vec{c} \times \vec{a} = (2y - z)\hat{i} - (2x - z)\hat{j} + (y - x)\hat{k}
\]
### Step 4: Calculate the Scalar Triple Product
Now we need to compute the scalar triple product \([\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}]\).
Using the formula:
\[
[\vec{a}, \vec{b}, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})
\]
We can express the scalar triple product as:
\[
[\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}] = \vec{a} \cdot (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a})
\]
### Step 5: Maximizing the Value
To maximize the value of the scalar triple product, we need to find the conditions under which \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are either parallel or anti-parallel.
The maximum value occurs when the angle between the vectors is either \(0\) or \(180\) degrees, which leads to the maximum value of the dot product being equal to the product of their magnitudes.
### Final Calculation
The maximum value of the expression is:
\[
\text{Maximum Value} = \sqrt{5}^2 = 5
\]
Thus, the maximum value of \([ \vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a} ]\) is:
\[
\boxed{5}
\]
