If a, b and c are non - zero real numbers and if system of equations `(a-1)x=y+z, (b-1)y=z+x and (c-1)z=x+y` have a non - trivial solutin, then `(3)/(2a)+(3)/(2b)+(3)/(2c)` is equal to

Answer

To solve the problem step by step, we need to analyze the given system of equations and determine the condition for a non-trivial solution. The equations are:

1. \((a-1)x = y + z\)
2. \((b-1)y = z + x\)
3. \((c-1)z = x + y\)

We can rewrite these equations in standard form:

1. \((a-1)x - y - z = 0\)
2. \(-x + (b-1)y - z = 0\)
3. \(-x - y + (c-1)z = 0\)

### Step 1: Form the Coefficient Matrix
We can express the system in matrix form \(A\mathbf{X} = 0\), where \(\mathbf{X} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\) and \(A\) is the coefficient matrix:

\[ A = \begin{bmatrix} a-1 & -1 & -1 \\ -1 & b-1 & -1 \\ -1 & -1 & c-1 \end{bmatrix} \]

### Step 2: Set the Determinant to Zero
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:

\[ \text{det}(A) = 0 \]

### Step 3: Calculate the Determinant
We calculate the determinant of matrix \(A\):

\[ \text{det}(A) = (a-1)\left((b-1)(c-1) - 1\right) + 1\left(-1(c-1) + 1\right) - 1\left(-1(b-1) + 1\right) \]

Expanding this determinant gives:

\[ = (a-1)((b-1)(c-1) - 1) + (c-1) - (b-1) \]

### Step 4: Simplify the Determinant
After simplifying, we get:

\[ = (a-1)(bc - b - c + 1) + c - b \]

Setting this equal to zero gives us a condition involving \(a\), \(b\), and \(c\).

### Step 5: Rearranging the Condition
From the determinant condition, we can derive:

\[ abc - ab - ac - bc + a + b + c - 1 = 0 \]

Rearranging gives:

\[ abc = ab + ac + bc - a - b - c + 1 \]

### Step 6: Finding the Value of \(\frac{3}{2a} + \frac{3}{2b} + \frac{3}{2c}\)
From the derived condition, we can express:

\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 \]

Thus,

\[ \frac{3}{2a} + \frac{3}{2b} + \frac{3}{2c} = \frac{3}{2} \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = \frac{3}{2} \cdot 1 = \frac{3}{2} \]

### Final Answer
The value of \(\frac{3}{2a} + \frac{3}{2b} + \frac{3}{2c}\) is:

\[ \frac{3}{2} \]