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 will follow the given instructions and derive the necessary equations.

### Step 1: Set up the matrix equation
We have the matrix equation:
\[ \begin{pmatrix} \alpha & \beta & \gamma \end{pmatrix} \begin{pmatrix} 2 & -1 & 1 \\ -1 & -1 & -2 \\ -1 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} \]

### Step 2: Multiply the matrices
We will multiply the row matrix \((\alpha, \beta, \gamma)\) with each column of the given matrix:

1. First column:
\[ 2\alpha - \beta + \gamma = 0 \]

2. Second column:
\[ -\alpha - \beta + 2\gamma = 0 \]

3. Third column:
\[ -\alpha + 2\beta + \gamma = 0 \]

### Step 3: Write the equations
From the multiplication, we have the following system of equations:
1. \(2\alpha - \beta + \gamma = 0\) (Equation 1)
2. \(-\alpha - \beta + 2\gamma = 0\) (Equation 2)
3. \(-\alpha + 2\beta + \gamma = 0\) (Equation 3)

### Step 4: Solve the equations
We will solve these equations step by step.

**From Equation 1:**
\[ \gamma = -2\alpha + \beta \quad \text{(Rearranging Equation 1)} \]

**Substituting \(\gamma\) into Equation 2:**
\[ -\alpha - \beta + 2(-2\alpha + \beta) = 0 \]
\[ -\alpha - \beta - 4\alpha + 2\beta = 0 \]
\[ -5\alpha + \beta = 0 \quad \Rightarrow \quad \beta = 5\alpha \quad \text{(Equation 4)} \]

**Substituting \(\beta\) into Equation 1:**
\[ 2\alpha - (5\alpha) + \gamma = 0 \]
\[ -3\alpha + \gamma = 0 \quad \Rightarrow \quad \gamma = 3\alpha \quad \text{(Equation 5)} \]

### Step 5: Substitute into the plane equation
We know that point \(A(\alpha, \beta, \gamma)\) lies on the plane \(2x + y + 3z = 2\). Substituting \(\beta\) and \(\gamma\) from Equations 4 and 5:
\[ 2\alpha + 5\alpha + 3(3\alpha) = 2 \]
\[ 2\alpha + 5\alpha + 9\alpha = 2 \]
\[ 16\alpha = 2 \quad \Rightarrow \quad \alpha = \frac{1}{8} \]

### Step 6: Find \(\beta\) and \(\gamma\)
Using \(\alpha\) to find \(\beta\) and \(\gamma\):
\[ \beta = 5\alpha = 5 \times \frac{1}{8} = \frac{5}{8} \]
\[ \gamma = 3\alpha = 3 \times \frac{1}{8} = \frac{3}{8} \]

### Step 7: Calculate \(3\alpha + 3\beta - 6\gamma\)
Now we can find \(3\alpha + 3\beta - 6\gamma\):
\[ 3\alpha + 3\beta - 6\gamma = 3\left(\frac{1}{8}\right) + 3\left(\frac{5}{8}\right) - 6\left(\frac{3}{8}\right) \]
\[ = \frac{3}{8} + \frac{15}{8} - \frac{18}{8} \]
\[ = \frac{3 + 15 - 18}{8} = \frac{0}{8} = 0 \]

### Final Answer
Thus, \(3\alpha + 3\beta - 6\gamma = 0\).

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