Let a,b, and c be three real numbers satistying `[a,b,c][(1,9,7),(8,2,7),(7,3,7)]=[0,0,0]` If the point `P(a,b,c)` with reference to (E), lies on the plane `2x+y+z=1`, the the value of `7a+b+c` is (A) 0 (B) 12 (C) 7 (D) 6

Options

A. 0
B. 12
C. 7
D. 6

Explanation

To solve the problem step by step, we start with the given matrix equation and the plane equation.

### Step 1: Set up the matrix equation
We have the equation:
\[ [a, b, c] \begin{pmatrix} 1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7 \end{pmatrix} = [0, 0, 0] \]
This implies that:
1. \( a + 8b + 7c = 0 \) (Equation 1)
2. \( 9a + 2b + 3c = 0 \) (Equation 2)
3. \( 7a + 7b + 7c = 0 \) (Equation 3)

### Step 2: Simplify Equation 3
From Equation 3, we can factor out 7:
\[ 7(a + b + c) = 0 \implies a + b + c = 0 \implies c = -a - b \quad \text{(Equation 4)} \]

### Step 3: Substitute Equation 4 into Equations 1 and 2
Now we substitute \( c = -a - b \) into Equations 1 and 2.

**Substituting into Equation 1:**
\[ a + 8b + 7(-a - b) = 0 \]
This simplifies to:
\[ a + 8b - 7a - 7b = 0 \implies -6a + b = 0 \implies b = 6a \quad \text{(Equation 5)} \]

**Substituting into Equation 2:**
\[ 9a + 2b + 3(-a - b) = 0 \]
This simplifies to:
\[ 9a + 2b - 3a - 3b = 0 \implies 6a - b = 0 \implies b = 6a \quad \text{(consistent with Equation 5)} \]

### Step 4: Substitute \( b = 6a \) back into Equation 4
Now substitute \( b = 6a \) into Equation 4:
\[ c = -a - b = -a - 6a = -7a \quad \text{(Equation 6)} \]

### Step 5: Use the plane equation
The point \( P(a, b, c) \) lies on the plane given by:
\[ 2x + y + z = 1 \]
Substituting \( x = a \), \( y = 6a \), and \( z = -7a \):
\[ 2a + 6a - 7a = 1 \]
This simplifies to:
\[ a = 1 \]

### Step 6: Find values of \( b \) and \( c \)
Using \( a = 1 \):
\[ b = 6a = 6 \times 1 = 6 \]
\[ c = -7a = -7 \times 1 = -7 \]

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

Thus, the value of \( 7a + b + c \) is:
\[ \boxed{6} \]