If `x =a, y=b, z=c` is a solution of the system of linear equations x + 8y + 7z =0, 9 x + 2y + 3z =0, x + y+z=0 such that point `(a,b,c )` lies on the plane `x + 2y + z=6,` then `2a+ b+c` equals

Options

A. (a) 1
B. (b) 2
C. (c) `-1`
D. (d) 0

Answer

To solve the problem step by step, we will analyze the given equations and find the values of \( a \), \( b \), and \( c \) based on the conditions provided.

### Step 1: Write down the equations
We have the following system of linear equations:
1. \( x + 8y + 7z = 0 \) (Equation 1)
2. \( 9x + 2y + 3z = 0 \) (Equation 2)
3. \( x + y + z = 0 \) (Equation 3)

### Step 2: Simplify the equations
To eliminate one variable, we can manipulate these equations. Let's multiply Equation 3 by 3:
\[ 3(x + y + z) = 3(0) \implies 3x + 3y + 3z = 0 \quad \text{(Equation 4)} \]

### Step 3: Subtract Equation 4 from Equation 2
Now, we will subtract Equation 4 from Equation 2:
\[ (9x + 2y + 3z) - (3x + 3y + 3z) = 0 \]
This simplifies to:
\[ 6x - y = 0 \implies y = 6x \quad \text{(Equation 5)} \]

### Step 4: Substitute \( y \) into Equation 1
Now we substitute \( y = 6x \) into Equation 1:
\[ x + 8(6x) + 7z = 0 \]
This simplifies to:
\[ x + 48x + 7z = 0 \implies 49x + 7z = 0 \]
From this, we can express \( z \) in terms of \( x \):
\[ 7z = -49x \implies z = -7x \quad \text{(Equation 6)} \]

### Step 5: Relate \( a \), \( b \), and \( c \)
From the equations derived:
- \( x = a \)
- \( y = b = 6a \)
- \( z = c = -7a \)

### Step 6: Substitute into the plane equation
The point \( (a, b, c) \) lies on the plane given by:
\[ x + 2y + z = 6 \]
Substituting \( a \), \( b \), and \( c \):
\[ a + 2(6a) + (-7a) = 6 \]
This simplifies to:
\[ a + 12a - 7a = 6 \implies 6a = 6 \]
Thus, we find:
\[ a = 1 \]

### Step 7: Find \( b \) and \( c \)
Now substituting \( a = 1 \) into the equations for \( b \) and \( c \):
\[ b = 6a = 6(1) = 6 \]
\[ c = -7a = -7(1) = -7 \]

### Step 8: Calculate \( 2a + b + c \)
Now we compute:
\[ 2a + b + c = 2(1) + 6 + (-7) = 2 + 6 - 7 = 1 \]

### Final Answer
Thus, the value of \( 2a + b + c \) is \( \boxed{1} \).