If `veca=2hati + hatj + hatk`, `vecb=hati + 2hatj + 2hatk` then `[veca vecb veci] hati` + `[veca vecb vecj] hatj` + `[veca vecb hatk] k` is equal to

Explanation

To solve the problem, we need to calculate the expression given by:

\[ [\vec{a} \, \vec{b} \, \vec{i}] \hat{i} + [\vec{a} \, \vec{b} \, \vec{j}] \hat{j} + [\vec{a} \, \vec{b} \, \hat{k}] \hat{k} \]

where \(\vec{a} = 2\hat{i} + \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}\).

### Step 1: Calculate \([\vec{a} \, \vec{b} \, \vec{i}]\)

We set up the determinant:

\[ [\vec{a} \, \vec{b} \, \vec{i}] = \begin{vmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 0 \end{vmatrix} \]

Calculating this determinant:

\[ = 2 \begin{vmatrix} 2 & 0 \\ 0 & 0 \end{vmatrix} - 1 \begin{vmatrix} 1 & 0 \\ 1 & 0 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 1 & 0 \end{vmatrix} \]

Calculating the smaller determinants:

\[ = 2(0) - 1(0) + 1(0 - 2) = 0 - 0 - 2 = -2 \]

So, \([\vec{a} \, \vec{b} \, \vec{i}] = -2\).

### Step 2: Calculate \([\vec{a} \, \vec{b} \, \vec{j}]\)

Now we calculate:

\[ [\vec{a} \, \vec{b} \, \vec{j}] = \begin{vmatrix} 2 & 1 & 1 \\ 1 & 2 & 2 \\ 0 & 1 & 0 \end{vmatrix} \]

Calculating this determinant:

\[ = 2 \begin{vmatrix} 2 & 2 \\ 1 & 0 \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ 0 & 0 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} \]

Calculating the smaller determinants:

\[ = 2(0 - 2) - 1(0) + 1(1 - 0) = -4 + 0 + 1 = -3 \]

So, \([\vec{a} \, \vec{b} \, \vec{j}] = -3\).

### Step 3: Calculate \([\vec{a} \, \vec{b} \, \hat{k}]\)

Now we calculate:

\[ [\vec{a} \, \vec{b} \, \hat{k}] = \begin{vmatrix} 2 & 1 & 1 \\ 1 & 2 & 2 \\ 0 & 0 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ = 2 \begin{vmatrix} 2 & 2 \\ 0 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 0 & 0 \end{vmatrix} \]

Calculating the smaller determinants:

\[ = 2(2) - 1(1) + 1(0) = 4 - 1 + 0 = 3 \]

So, \([\vec{a} \, \vec{b} \, \hat{k}] = 3\).

### Step 4: Combine the Results

Now we can substitute the values we found back into the original expression:

\[ [\vec{a} \, \vec{b} \, \vec{i}] \hat{i} + [\vec{a} \, \vec{b} \, \vec{j}] \hat{j} + [\vec{a} \, \vec{b} \, \hat{k}] \hat{k} \]

This becomes:

\[ (-2) \hat{i} + (-3) \hat{j} + (3) \hat{k} \]

Thus, the final result is:

\[ -2 \hat{i} - 3 \hat{j} + 3 \hat{k} \]

### Final Answer:

\[ \boxed{-2 \hat{i} - 3 \hat{j} + 3 \hat{k}} \]