The value of `a` for which the image of the point `(a, a-1)` w.r.t the line mirror `3x+y=6a` is the point `(a^2 + 1, a)` is (A) 0 (B) 1 (C) 2 (D) none of these

Explanation

To solve the problem, we need to find the value of \( a \) such that the image of the point \( (a, a-1) \) with respect to the line \( 3x + y = 6a \) is the point \( (a^2 + 1, a) \).

### Step-by-Step Solution:

1. **Identify the given points and line**:
- Point \( P = (a, a-1) \)
- Image point \( P' = (a^2 + 1, a) \)
- Line equation: \( 3x + y = 6a \)

2. **Find the midpoint of \( P \) and \( P' \)**:
The midpoint \( M \) of points \( P \) and \( P' \) is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{a + (a^2 + 1)}{2}, \frac{(a - 1) + a}{2} \right) \]
Simplifying this gives:
\[ M = \left( \frac{a^2 + a + 1}{2}, \frac{2a - 1}{2} \right) \]

3. **Substitute the midpoint into the line equation**:
Since the midpoint \( M \) lies on the line \( 3x + y = 6a \), we substitute \( M \) into the line equation:
\[ 3\left( \frac{a^2 + a + 1}{2} \right) + \left( \frac{2a - 1}{2} \right) = 6a \]
Multiplying through by 2 to eliminate the fractions:
\[ 3(a^2 + a + 1) + (2a - 1) = 12a \]
Expanding this gives:
\[ 3a^2 + 3a + 3 + 2a - 1 = 12a \]
Simplifying further:
\[ 3a^2 + 5a + 2 = 12a \]
Rearranging the equation:
\[ 3a^2 - 7a + 2 = 0 \]

4. **Solve the quadratic equation**:
We can solve the quadratic equation \( 3a^2 - 7a + 2 = 0 \) using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 3 \), \( b = -7 \), and \( c = 2 \):
\[ a = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = \frac{7 \pm \sqrt{49 - 24}}{6} = \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6} \]
This gives us two possible solutions:
\[ a = \frac{12}{6} = 2 \quad \text{and} \quad a = \frac{2}{6} = \frac{1}{3} \]

5. **Check the options**:
The options given are \( 0, 1, 2, \) and \( \text{none of these} \). Among the solutions \( a = 2 \) and \( a = \frac{1}{3} \), only \( a = 2 \) is an option.

### Final Answer:
The value of \( a \) for which the image of the point \( (a, a-1) \) with respect to the line \( 3x + y = 6a \) is the point \( (a^2 + 1, a) \) is \( \boxed{2} \).