For any real `a, b, c` the expression `2a^(2) + 11b^(2) + 4c^(2) - 8ab - 6bc - 2c + 41` is minimum then

Options

A. `a - b - c = 0`
B. `a = 2b`
C. `a + b + c = 4`
D. `c = a`

Explanation

To find the minimum value of the expression \(2a^2 + 11b^2 + 4c^2 - 8ab - 6bc - 2c + 41\), we will complete the square for the quadratic terms in \(a\), \(b\), and \(c\).

### Step 1: Rearranging the Expression
We start with the expression:
\[ 2a^2 + 11b^2 + 4c^2 - 8ab - 6bc - 2c + 41 \]

### Step 2: Grouping Terms
We can group the terms involving \(a\) and \(b\) together, and the terms involving \(b\) and \(c\) together:
\[ = 2a^2 - 8ab + 11b^2 + 4c^2 - 6bc - 2c + 41 \]

### Step 3: Completing the Square for \(a\) and \(b\)
To complete the square for \(2a^2 - 8ab + 11b^2\):
1. Factor out the coefficient of \(a^2\):
\[ = 2(a^2 - 4ab) + 11b^2 \]
2. Complete the square:
\[ = 2((a - 2b)^2 - 4b^2) + 11b^2 = 2(a - 2b)^2 + (11b^2 - 8b^2) = 2(a - 2b)^2 + 3b^2 \]

### Step 4: Completing the Square for \(b\) and \(c\)
Next, we complete the square for \(3b^2 - 6bc + 4c^2\):
1. Factor out the coefficient of \(b^2\):
\[ = 3(b^2 - 2bc) + 4c^2 \]
2. Complete the square:
\[ = 3((b - c)^2 - c^2) + 4c^2 = 3(b - c)^2 + (4c^2 - 3c^2) = 3(b - c)^2 + c^2 \]

### Step 5: Putting it All Together
Now, substitute back into the expression:
\[ = 2(a - 2b)^2 + 3(b - c)^2 + c^2 + 41 \]

### Step 6: Finding the Minimum Value
The expression \(2(a - 2b)^2 + 3(b - c)^2 + c^2\) is minimized when each squared term is zero:
1. \(a - 2b = 0 \Rightarrow a = 2b\)
2. \(b - c = 0 \Rightarrow b = c\)
3. \(c = 0\)

### Step 7: Solving for \(a\), \(b\), and \(c\)
From \(b = c\) and substituting into \(a = 2b\):
1. Let \(c = 1\), then \(b = 1\) and \(a = 2\).

### Conclusion
The minimum value of the expression occurs when:
\[ a = 2, \quad b = 1, \quad c = 1 \]
The minimum value of the expression is \(41\).