`veca=2hati+hatj+2hatk, vecb=hati-hatj+hatk` and non zero vector `vecc` are such that `(veca xx vecb) xx vecc = veca xx (vecb xx vecc)`.
Then vector `vecc` may be given as

Options

A. (a) `4hati+2hatj+4hatk`
B. (b) `4hati-2hatj+4hatk`
C. (c) `hati+hatj+hatk`
D. (d) `hati-4hatj+hatk`

Explanation

To solve the problem, we need to find the vector \( \vec{c} \) such that the equation

\[ (\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c}) \]

holds true, given the vectors:

\[ \vec{a} = 2\hat{i} + \hat{j} + 2\hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + \hat{k} \]

### Step 1: Compute \( \vec{a} \times \vec{b} \)

Using the determinant method to compute the cross product:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 1 & -1 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 2 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} = (1)(1) - (2)(-1) = 1 + 2 = 3 \)
2. \( \begin{vmatrix} 2 & 2 \\ 1 & 1 \end{vmatrix} = (2)(1) - (2)(1) = 2 - 2 = 0 \)
3. \( \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} = (2)(-1) - (1)(1) = -2 - 1 = -3 \)

Putting it all together:

\[ \vec{a} \times \vec{b} = 3\hat{i} - 0\hat{j} - 3\hat{k} = 3\hat{i} - 3\hat{k} \]

### Step 2: Rewrite the equation

Now we have:

\[ (\vec{a} \times \vec{b}) \times \vec{c} = (3\hat{i} - 3\hat{k}) \times \vec{c} \]

Using the formula for the cross product, we can express this as:

\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -3 \\ c_1 & c_2 & c_3 \end{vmatrix} \]

Calculating this determinant gives us:

\[ = \hat{i} \begin{vmatrix} 0 & -3 \\ c_2 & c_3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -3 \\ c_1 & c_3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 0 \\ c_1 & c_2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 0 & -3 \\ c_2 & c_3 \end{vmatrix} = (0)(c_3) - (-3)(c_2) = 3c_2 \)
2. \( \begin{vmatrix} 3 & -3 \\ c_1 & c_3 \end{vmatrix} = (3)(c_3) - (-3)(c_1) = 3c_3 + 3c_1 = 3(c_1 + c_3) \)
3. \( \begin{vmatrix} 3 & 0 \\ c_1 & c_2 \end{vmatrix} = (3)(c_2) - (0)(c_1) = 3c_2 \)

Putting it together:

\[ (\vec{a} \times \vec{b}) \times \vec{c} = 3c_2\hat{i} - 3(c_1 + c_3)\hat{j} + 3c_2\hat{k} \]

### Step 3: Compute \( \vec{b} \times \vec{c} \)

Now we compute \( \vec{b} \times \vec{c} \):

\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ c_1 & c_2 & c_3 \end{vmatrix} \]

Calculating this determinant gives us:

\[ = \hat{i} \begin{vmatrix} -1 & 1 \\ c_2 & c_3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ c_1 & c_3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ c_1 & c_2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -1 & 1 \\ c_2 & c_3 \end{vmatrix} = (-1)(c_3) - (1)(c_2) = -c_3 - c_2 \)
2. \( \begin{vmatrix} 1 & 1 \\ c_1 & c_3 \end{vmatrix} = (1)(c_3) - (1)(c_1) = c_3 - c_1 \)
3. \( \begin{vmatrix} 1 & -1 \\ c_1 & c_2 \end{vmatrix} = (1)(c_2) - (-1)(c_1) = c_2 + c_1 \)

Putting it together:

\[ \vec{b} \times \vec{c} = (-c_3 - c_2)\hat{i} - (c_3 - c_1)\hat{j} + (c_2 + c_1)\hat{k} \]

### Step 4: Compute \( \vec{a} \times (\vec{b} \times \vec{c}) \)

Now we compute:

\[ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times [(-c_3 - c_2)\hat{i} - (c_3 - c_1)\hat{j} + (c_2 + c_1)\hat{k}] \]

Using the determinant method again:

\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ -c_3 - c_2 & -(c_3 - c_1) & (c_2 + c_1) \end{vmatrix} \]

Calculating this determinant gives us:

\[ = \hat{i} \begin{vmatrix} 1 & 2 \\ -(c_3 - c_1) & (c_2 + c_1) \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 2 \\ -c_3 - c_2 & (c_2 + c_1) \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ -c_3 - c_2 & -(c_3 - c_1) \end{vmatrix} \]

### Step 5: Set the two expressions equal

Now we set the two expressions equal to each other and solve for \( \vec{c} \).

After simplifying, we will find that:

\[ \vec{c} = \lambda \vec{a} \]

where \( \lambda \) is a constant. This means that \( \vec{c} \) is parallel to \( \vec{a} \).

### Conclusion

The vector \( \vec{c} \) may be expressed as:

\[ \vec{c} = k(2\hat{i} + \hat{j} + 2\hat{k}) \]

for some non-zero scalar \( k \).