The vectors which is/are coplanar with vectors `hati+hatj+2hatk and hati+2hatj+hatk` and perpendicular to the vector `hati+hatj+hatk ` is /are (A) `hatj-hatk` (B) `-hati+hatj` (C) `hati-hatj` (D) `-hatj+hatk`

Options

A. `(2hati + hatj - 3hatk)/(sqrt14)`
B. `(hatj + hatk)/(sqrt2)`
C. `(hati - hatj)/(sqrt(2))`
D. `(hati - hatk)/(sqrt(2))`

Explanation

To solve the problem, we need to find the vectors that are coplanar with the vectors \( \mathbf{v_1} = \hat{i} + \hat{j} + 2\hat{k} \) and \( \mathbf{v_2} = \hat{i} + \hat{j} + \hat{k} \), and also perpendicular to the vector \( \mathbf{v_3} = \hat{i} + \hat{j} + \hat{k} \).

### Step 1: Determine the condition for coplanarity
Vectors \( \mathbf{v} \), \( \mathbf{v_1} \), and \( \mathbf{v_2} \) are coplanar if the scalar triple product is zero. This can be expressed using the determinant of a matrix formed by these vectors:

\[ \begin{vmatrix} a & b & c \\ 1 & 1 & 2 \\ 1 & 1 & 1 \end{vmatrix} = 0 \]

Where \( \mathbf{v} = a\hat{i} + b\hat{j} + c\hat{k} \).

### Step 2: Calculate the determinant
Calculating the determinant gives us:

\[ a \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} - b \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} + c \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} \]

Calculating the smaller determinants:

\[ \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (2)(1) = 1 - 2 = -1 \]

\[ \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(1) = 0 \]

Thus, the determinant simplifies to:

\[ a(-1) - b(-1) + c(0) = 0 \implies -a + b = 0 \implies b = a \]

### Step 3: Determine the condition for perpendicularity
For \( \mathbf{v} \) to be perpendicular to \( \mathbf{v_3} \), we have:

\[ \mathbf{v} \cdot \mathbf{v_3} = 0 \]

This gives us:

\[ a + b + c = 0 \]

### Step 4: Substitute \( b = a \) into the perpendicularity condition
Substituting \( b = a \) into the equation \( a + b + c = 0 \):

\[ a + a + c = 0 \implies 2a + c = 0 \implies c = -2a \]

### Step 5: Express the vector \( \mathbf{v} \)
Now we can express \( \mathbf{v} \):

\[ \mathbf{v} = a\hat{i} + a\hat{j} - 2a\hat{k} = a(\hat{i} + \hat{j} - 2\hat{k}) \]

### Step 6: Check the options
Now we need to check which of the given options can be expressed in the form \( a(\hat{i} + \hat{j} - 2\hat{k}) \):

- (A) \( \hat{j} - \hat{k} \) can be written as \( 0\hat{i} + 1\hat{j} - 1\hat{k} \) (not in the required form).
- (B) \( -\hat{i} + \hat{j} \) can be written as \( -1\hat{i} + 1\hat{j} + 0\hat{k} \) (not in the required form).
- (C) \( \hat{i} - \hat{j} \) can be written as \( 1\hat{i} - 1\hat{j} + 0\hat{k} \) (not in the required form).
- (D) \( -\hat{j} + \hat{k} \) can be written as \( 0\hat{i} - 1\hat{j} + 1\hat{k} \) (not in the required form).

### Conclusion
The only vectors that satisfy both conditions are:

- \( \hat{j} - \hat{k} \) (Option A)
- \( -\hat{j} + \hat{k} \) (Option D)

Thus, the correct options are (A) and (D).