Let a, b and c satisfy the system of equations
`a+2b+3c=6, 4a+5b+6c=12 " rad "6a+9b=4`. If the roots of the equation `(a+b+c)x^(2)-abcx` are `alpha` and `beta+(a^(-1)+b^(-1)+c^(-1))=0` then `(1)/(alpha)+(1)/(beta)` is equal to

Options

A. 243
B. 100
C. `(243)/(12)`
D. `(100)/(243)`

Answer

To solve the problem step by step, we will follow the outlined approach.

### Step 1: Write the system of equations
We have the following system of equations:
1. \( a + 2b + 3c = 6 \) (Equation 1)
2. \( 4a + 5b + 6c = 12 \) (Equation 2)
3. \( 6a + 9b = 4 \) (Equation 3)

### Step 2: Set up the determinant (Δ)
We will use Cramer's rule to solve for \( a \), \( b \), and \( c \). First, we need to set up the determinant \( Δ \) using the coefficients of \( a \), \( b \), and \( c \).

\[ Δ = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 6 & 9 & 0 \end{vmatrix} \]

### Step 3: Calculate the determinant (Δ)
Using the determinant formula, we calculate:

\[ Δ = 1 \cdot \begin{vmatrix} 5 & 6 \\ 9 & 0 \end{vmatrix} - 2 \cdot \begin{vmatrix} 4 & 6 \\ 6 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & 5 \\ 6 & 9 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 5 & 6 \\ 9 & 0 \end{vmatrix} = (5 \cdot 0) - (6 \cdot 9) = -54 \)
2. \( \begin{vmatrix} 4 & 6 \\ 6 & 0 \end{vmatrix} = (4 \cdot 0) - (6 \cdot 6) = -36 \)
3. \( \begin{vmatrix} 4 & 5 \\ 6 & 9 \end{vmatrix} = (4 \cdot 9) - (5 \cdot 6) = 36 - 30 = 6 \)

Now substituting back into the determinant:

\[ Δ = 1 \cdot (-54) - 2 \cdot (-36) + 3 \cdot 6 = -54 + 72 + 18 = 36 \]

### Step 4: Calculate \( Δ_1 \), \( Δ_2 \), and \( Δ_3 \)
Next, we calculate \( Δ_1 \), \( Δ_2 \), and \( Δ_3 \):

- **For \( Δ_1 \)** (replace the first column with the RHS):
\[ Δ_1 = \begin{vmatrix} 6 & 2 & 3 \\ 12 & 5 & 6 \\ 4 & 9 & 0 \end{vmatrix} \]
Calculating \( Δ_1 \):
\[ Δ_1 = 6 \cdot \begin{vmatrix} 5 & 6 \\ 9 & 0 \end{vmatrix} - 2 \cdot \begin{vmatrix} 12 & 6 \\ 4 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 12 & 5 \\ 4 & 9 \end{vmatrix} \]
Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 5 & 6 \\ 9 & 0 \end{vmatrix} = -54 \)
2. \( \begin{vmatrix} 12 & 6 \\ 4 & 0 \end{vmatrix} = -24 \)
3. \( \begin{vmatrix} 12 & 5 \\ 4 & 9 \end{vmatrix} = 108 - 20 = 88 \)

Now substituting back into \( Δ_1 \):
\[ Δ_1 = 6 \cdot (-54) - 2 \cdot (-24) + 3 \cdot 88 = -324 + 48 + 264 = -12 \]

- **For \( Δ_2 \)** (replace the second column with the RHS):
\[ Δ_2 = \begin{vmatrix} 1 & 6 & 3 \\ 4 & 12 & 6 \\ 6 & 4 & 0 \end{vmatrix} \]
Calculating \( Δ_2 \):
\[ Δ_2 = 1 \cdot \begin{vmatrix} 12 & 6 \\ 4 & 0 \end{vmatrix} - 6 \cdot \begin{vmatrix} 4 & 6 \\ 6 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & 12 \\ 6 & 4 \end{vmatrix} \]
Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 12 & 6 \\ 4 & 0 \end{vmatrix} = -24 \)
2. \( \begin{vmatrix} 4 & 6 \\ 6 & 0 \end{vmatrix} = -36 \)
3. \( \begin{vmatrix} 4 & 12 \\ 6 & 4 \end{vmatrix} = 16 - 72 = -56 \)

Now substituting back into \( Δ_2 \):
\[ Δ_2 = 1 \cdot (-24) - 6 \cdot (-36) + 3 \cdot (-56) = -24 + 216 - 168 = 24 \]

- **For \( Δ_3 \)** (replace the third column with the RHS):
\[ Δ_3 = \begin{vmatrix} 1 & 2 & 6 \\ 4 & 5 & 12 \\ 6 & 9 & 4 \end{vmatrix} \]
Calculating \( Δ_3 \):
\[ Δ_3 = 1 \cdot \begin{vmatrix} 5 & 12 \\ 9 & 4 \end{vmatrix} - 2 \cdot \begin{vmatrix} 4 & 12 \\ 6 & 4 \end{vmatrix} + 6 \cdot \begin{vmatrix} 4 & 5 \\ 6 & 9 \end{vmatrix} \]
Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 5 & 12 \\ 9 & 4 \end{vmatrix} = 20 - 108 = -88 \)
2. \( \begin{vmatrix} 4 & 12 \\ 6 & 4 \end{vmatrix} = -56 \)
3. \( \begin{vmatrix} 4 & 5 \\ 6 & 9 \end{vmatrix} = 36 - 30 = 6 \)

Now substituting back into \( Δ_3 \):
\[ Δ_3 = 1 \cdot (-88) - 2 \cdot (-56) + 6 \cdot 6 = -88 + 112 + 36 = 60 \]

### Step 5: Calculate \( a \), \( b \), and \( c \)
Using Cramer's rule:
\[ a = \frac{Δ_1}{Δ} = \frac{-12}{36} = -\frac{1}{3} \]
\[ b = \frac{Δ_2}{Δ} = \frac{24}{36} = \frac{2}{3} \]
\[ c = \frac{Δ_3}{Δ} = \frac{60}{36} = \frac{5}{3} \]

### Step 6: Form the quadratic equation
The quadratic equation is given by:
\[ (a + b + c)x^2 - abc x = 0 \]
Calculating \( a + b + c \):
\[ a + b + c = -\frac{1}{3} + \frac{2}{3} + \frac{5}{3} = \frac{6}{3} = 2 \]
Calculating \( abc \):
\[ abc = -\frac{1}{3} \cdot \frac{2}{3} \cdot \frac{5}{3} = -\frac{10}{27} \]
Thus, the quadratic equation becomes:
\[ 2x^2 + \frac{10}{27}x = 0 \]
Multiplying through by 27 to eliminate the fraction:
\[ 54x^2 + 10x = 0 \]

### Step 7: Find the roots
Factoring out \( x \):
\[ x(54x + 10) = 0 \]
The roots are:
\[ x = 0 \quad \text{or} \quad 54x + 10 = 0 \Rightarrow x = -\frac{10}{54} = -\frac{5}{27} \]
Thus, the roots are \( \alpha = 0 \) and \( \beta = -\frac{5}{27} \).

### Step 8: Calculate \( \frac{1}{\alpha} + \frac{1}{\beta} \)
We need to find:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{0} + \frac{1}{-\frac{5}{27}} \]
Since \( \frac{1}{0} \) is undefined, we can only consider the second term:
\[ \frac{1}{\beta} = -\frac{27}{5} \]

### Final Answer
Thus, the final answer is:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = -\frac{27}{5} \]