Consider the system of equations
`(a -1) x -y -z = 0`
`x -(b -1) y +z = 0`
`x + y - (c -1) z = 0`
Where a, b and c are non-zero real number
Statement1 : If x,y,z are not all zero, then `ab + bc + ca = abc`
Statement 2 : `abc ge 27`

Options

A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 3
B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 3
C. Statement 1 is true, Statement 2 is False
D. Statement 1 is False, Statement 2 is true

Explanation

To solve the given system of equations and analyze the statements, we will follow these steps:

### Step 1: Write the system of equations in matrix form
The system of equations is:
1. \((a - 1)x - y - z = 0\)
2. \(x - (b - 1)y + z = 0\)
3. \(x + y - (c - 1)z = 0\)

We can represent this system in matrix form as:
\[ \begin{bmatrix} a - 1 & -1 & -1 \\ 1 & -(b - 1) & 1 \\ 1 & 1 & -(c - 1) \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Set up the determinant
For the system to have a non-trivial solution (i.e., not all \(x, y, z\) are zero), the determinant of the coefficient matrix must be zero:
\[ \text{Det} = \begin{vmatrix} a - 1 & -1 & -1 \\ 1 & -(b - 1) & 1 \\ 1 & 1 & -(c - 1) \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
We will calculate the determinant using cofactor expansion. Expanding along the first row:
\[ \text{Det} = (a - 1) \begin{vmatrix} -(b - 1) & 1 \\ 1 & -(c - 1) \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ 1 & -(c - 1) \end{vmatrix} - 1 \begin{vmatrix} 1 & -(b - 1) \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \(\begin{vmatrix} -(b - 1) & 1 \\ 1 & -(c - 1) \end{vmatrix} = -(b - 1)(-(c - 1)) - 1 = (b - 1)(c - 1) - 1\)

2. \(\begin{vmatrix} 1 & 1 \\ 1 & -(c - 1) \end{vmatrix} = 1 \cdot (-(c - 1)) - 1 \cdot 1 = -(c - 1) - 1 = -c\)

3. \(\begin{vmatrix} 1 & -(b - 1) \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - 1 \cdot (-(b - 1)) = 1 + (b - 1) = b\)

Putting it all together:
\[ \text{Det} = (a - 1)((b - 1)(c - 1) - 1) - c - b \]

### Step 4: Set the determinant to zero
Setting the determinant to zero gives:
\[ (a - 1)((b - 1)(c - 1) - 1) - c - b = 0 \]

### Step 5: Rearranging the equation
After simplifying the determinant equation, we get:
\[ abc - ab - ac - bc + a + b + c - 1 = 0 \]
This can be rearranged to:
\[ ab + ac + bc = abc \]

### Step 6: Analyze Statement 1
From the above equation, we conclude that Statement 1 is true:
- If \(x, y, z\) are not all zero, then \(ab + ac + bc = abc\).

### Step 7: Analyze Statement 2
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{ab + ac + bc}{3} \geq \sqrt[3]{(abc)^2} \]
Given \(ab + ac + bc = abc\), we have:
\[ \frac{abc}{3} \geq \sqrt[3]{(abc)^2} \]
Cubing both sides leads to:
\[ \frac{(abc)^3}{27} \geq (abc)^2 \]
This implies:
\[ abc \geq 27 \]
Thus, Statement 2 is also true.

### Conclusion
Both statements are true. Therefore, the final answer is:
- **Both Statement 1 and Statement 2 are true.**

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