If the vectors `2ahati+bhatj+chatk, bhati+chatj+2ahatk` and `chati+2ahatj+bhatk` are coplanar vectors, then the straight lines `ax+by+c=0` will always pass through the point

Options

A. `(1,2)`
B. `(2,-1)`
C. `(2,1)`
D. `(1,-2)`

Explanation

To solve the problem, we need to determine the conditions under which the given vectors are coplanar and find the point through which the line \( ax + by + c = 0 \) passes.

### Step 1: Define the Vectors
Let:
- \( \mathbf{A} = 2a \hat{i} + b \hat{j} + c \hat{k} \)
- \( \mathbf{B} = b \hat{i} + c \hat{j} + 2a \hat{k} \)
- \( \mathbf{C} = c \hat{i} + 2a \hat{j} + b \hat{k} \)

### Step 2: Use the Condition for Coplanarity
The vectors \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) are coplanar if the scalar triple product is zero:
\[ \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 0 \]

### Step 3: Calculate the Cross Product \( \mathbf{B} \times \mathbf{C} \)
To find \( \mathbf{B} \times \mathbf{C} \), we can use the determinant of a matrix:
\[ \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ b & c & 2a \\ c & 2a & b \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \left( c \cdot b - 2a \cdot 2a \right) - \hat{j} \left( b \cdot b - c \cdot 2a \right) + \hat{k} \left( b \cdot 2a - c \cdot c \right) \]
\[ = \hat{i} (bc - 4a^2) - \hat{j} (b^2 - 2ac) + \hat{k} (2ab - c^2) \]

### Step 4: Calculate the Dot Product \( \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \)
Now we compute:
\[ \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = (2a \hat{i} + b \hat{j} + c \hat{k}) \cdot \left( (bc - 4a^2) \hat{i} - (b^2 - 2ac) \hat{j} + (2ab - c^2) \hat{k} \right) \]
\[ = 2a(bc - 4a^2) - b(b^2 - 2ac) + c(2ab - c^2) \]

### Step 5: Set the Expression to Zero
Setting the expression to zero for coplanarity:
\[ 2a(bc - 4a^2) - b(b^2 - 2ac) + c(2ab - c^2) = 0 \]

### Step 6: Simplify the Equation
Rearranging and simplifying the equation will yield a relationship between \( a, b, c \).

### Step 7: Find the Point of Intersection
From the condition derived, we can find that:
\[ 2a + b + c = 0 \]
This implies that the line \( ax + by + c = 0 \) passes through the point where \( x = 2 \), \( y = 1 \), and \( z = 1 \).

### Conclusion
The straight line \( ax + by + c = 0 \) will always pass through the point \( (2, 1) \).