Let `alpha,beta, gamma` be three real numbers satisfying `[(alpha,beta,gamma)][(2,-1,1),(-1,-1,-2),(-1,2,1)]=[(0,0,0)]`. If the point `A(alpha, beta, gamma)` lies on the plane `2x+y+3z=2`, then `3alpha+3beta-6gamma` is equal to

Options

A. 0
B. `-(1)/(3)`
C. 1
D. `-3`

Answer

To solve the problem step by step, we start with the given matrix equation:

\[ \begin{pmatrix} \alpha & \beta & \gamma \end{pmatrix} \begin{pmatrix} 2 & -1 & 1 \\ -1 & -1 & -2 \\ -1 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} \]

This implies that the following equations must hold:

1. \( 2\alpha - \beta + \gamma = 0 \) (from the first column)
2. \( -\alpha - \beta - 2\gamma = 0 \) (from the second column)
3. \( -\alpha + 2\beta + \gamma = 0 \) (from the third column)

Next, we can simplify these equations one by one.

### Step 1: Solve the first equation
From the first equation:
\[ 2\alpha - \beta + \gamma = 0 \implies \gamma = -2\alpha + \beta \]

### Step 2: Substitute into the second equation
Now, substitute \(\gamma\) into the second equation:
\[ -\alpha - \beta - 2(-2\alpha + \beta) = 0 \]
This simplifies to:
\[ -\alpha - \beta + 4\alpha - 2\beta = 0 \implies 3\alpha - 3\beta = 0 \implies \alpha = \beta \]

### Step 3: Substitute \(\alpha = \beta\) into the first equation
Now substitute \(\beta\) with \(\alpha\) in the first equation:
\[ 2\alpha - \alpha + \gamma = 0 \implies \gamma = -\alpha \]

### Step 4: Express \(\alpha, \beta, \gamma\)
Now we have:
\[ \beta = \alpha \quad \text{and} \quad \gamma = -\alpha \]

### Step 5: Use the plane equation
We know that point \(A(\alpha, \beta, \gamma)\) lies on the plane given by:
\[ 2x + y + 3z = 2 \]
Substituting \((\alpha, \alpha, -\alpha)\) into the plane equation:
\[ 2\alpha + \alpha + 3(-\alpha) = 2 \implies 2\alpha + \alpha - 3\alpha = 2 \implies 0 = 2 \]
This means that the equation holds true for any value of \(\alpha\).

### Step 6: Calculate \(3\alpha + 3\beta - 6\gamma\)
Now, we need to calculate:
\[ 3\alpha + 3\beta - 6\gamma \]
Substituting \(\beta = \alpha\) and \(\gamma = -\alpha\):
\[ 3\alpha + 3\alpha - 6(-\alpha) = 3\alpha + 3\alpha + 6\alpha = 12\alpha \]

### Step 7: Conclusion
Since \(\alpha\) can take any real value, \(3\alpha + 3\beta - 6\gamma\) is equal to \(12\alpha\).

Thus, the final answer is:
\[ \text{The value of } 3\alpha + 3\beta - 6\gamma \text{ is } 12\alpha. \]