f a point `P(alpha,beta,gamma)` satisfying `(alpha,beta,gamma)((2,10,8),(9,3,8),(8,4,8))=(0,0,0)` lies on the plane `2x+4y+3z=5,` then `6 alpha+9 beta+7 gamma` is equal to

Options

A. `(5)/(4)`
B. `-1`
C. `11`
D. `(11)/5`

Answer

To solve the problem, we need to find the value of \(6\alpha + 9\beta + 7\gamma\) given the conditions provided. Let's break down the solution step by step.

### Step 1: Set up the equations
We are given the point \(P(\alpha, \beta, \gamma)\) that satisfies the following matrix equation:

\[ (\alpha, \beta, \gamma) \begin{pmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{pmatrix} = (0, 0, 0) \]

This implies that:

1. \(2\alpha + 10\beta + 8\gamma = 0\) (Equation 1)
2. \(9\alpha + 3\beta + 8\gamma = 0\) (Equation 2)
3. \(8\alpha + 4\beta + 8\gamma = 0\) (Equation 3)

Additionally, the point lies on the plane defined by:

\[ 2x + 4y + 3z = 5 \]

Substituting \(x = \alpha\), \(y = \beta\), and \(z = \gamma\), we have:

\[ 2\alpha + 4\beta + 3\gamma = 5 \quad \text{(Equation 4)} \]

### Step 2: Solve the system of equations
Now we have four equations:

1. \(2\alpha + 10\beta + 8\gamma = 0\)
2. \(9\alpha + 3\beta + 8\gamma = 0\)
3. \(8\alpha + 4\beta + 8\gamma = 0\)
4. \(2\alpha + 4\beta + 3\gamma = 5\)

From Equation 1, we can express \(\gamma\) in terms of \(\alpha\) and \(\beta\):

\[ 8\gamma = -2\alpha - 10\beta \implies \gamma = -\frac{1}{4}\alpha - \frac{5}{4}\beta \quad \text{(Equation 5)} \]

### Step 3: Substitute \(\gamma\) in other equations
Substituting Equation 5 into Equations 2 and 3:

**For Equation 2:**

\[ 9\alpha + 3\beta + 8\left(-\frac{1}{4}\alpha - \frac{5}{4}\beta\right) = 0 \]
\[ 9\alpha + 3\beta - 2\alpha - 10\beta = 0 \]
\[ 7\alpha - 7\beta = 0 \implies \alpha = \beta \quad \text{(Equation 6)} \]

**For Equation 3:**

\[ 8\alpha + 4\beta + 8\left(-\frac{1}{4}\alpha - \frac{5}{4}\beta\right) = 0 \]
\[ 8\alpha + 4\beta - 2\alpha - 10\beta = 0 \]
\[ 6\alpha - 6\beta = 0 \implies \alpha = \beta \quad \text{(consistent with Equation 6)} \]

### Step 4: Substitute \(\beta\) in terms of \(\alpha\)
Now, substituting \(\beta = \alpha\) into Equation 5:

\[ \gamma = -\frac{1}{4}\alpha - \frac{5}{4}\alpha = -\frac{6}{4}\alpha = -\frac{3}{2}\alpha \quad \text{(Equation 7)} \]

### Step 5: Substitute into Equation 4
Now substitute \(\beta\) and \(\gamma\) into Equation 4:

\[ 2\alpha + 4\alpha + 3\left(-\frac{3}{2}\alpha\right) = 5 \]
\[ 2\alpha + 4\alpha - \frac{9}{2}\alpha = 5 \]
\[ \left(6 - \frac{9}{2}\right)\alpha = 5 \]
\[ \frac{12 - 9}{2}\alpha = 5 \implies \frac{3}{2}\alpha = 5 \implies \alpha = \frac{10}{3} \]

### Step 6: Find \(\beta\) and \(\gamma\)
Using \(\alpha = \frac{10}{3}\):

\[ \beta = \alpha = \frac{10}{3} \]
\[ \gamma = -\frac{3}{2}\alpha = -\frac{3}{2} \cdot \frac{10}{3} = -5 \]

### Step 7: Calculate \(6\alpha + 9\beta + 7\gamma\)
Now we calculate \(6\alpha + 9\beta + 7\gamma\):

\[ 6\alpha + 9\beta + 7\gamma = 6\left(\frac{10}{3}\right) + 9\left(\frac{10}{3}\right) + 7(-5) \]
\[ = \frac{60}{3} + \frac{90}{3} - 35 = 20 + 30 - 35 = 15 \]

Thus, the final answer is:

\[ \boxed{15} \]