If the system of equation ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 has non-trivial solution then find the value of `|{:(bc-a^2,ca-b^2,ab-c^2),(ca-b^2, ab-c^2, bc-a^2),(ab-c^2, bc-a^2, ca-b^2):}|`

Answer

To solve the problem, we need to find the value of the determinant

\[ D = \begin{vmatrix} bc - a^2 & ca - b^2 & ab - c^2 \\ ca - b^2 & ab - c^2 & bc - a^2 \\ ab - c^2 & bc - a^2 & ca - b^2 \end{vmatrix} \]

given that the system of equations has a non-trivial solution.

### Step 1: Understand the condition for non-trivial solutions
For the system of equations

\[ \begin{align*} ax + by + cz &= 0 \\ bx + cy + az &= 0 \\ cx + ay + bz &= 0 \end{align*} \]

to have a non-trivial solution, the determinant of the coefficients must be zero:

\[ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant
We can compute the determinant using the property of determinants. We can add all three rows together:

\[ R_1 + R_2 + R_3 \Rightarrow (a+b+c, a+b+c, a+b+c) \]

This gives us:

\[ \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ b & c & a \\ c & a & b \end{vmatrix} \]

### Step 3: Factor out the common term
Factoring out \(a+b+c\) from the first row, we have:

\[ (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix} \]

### Step 4: Calculate the new determinant
Now we need to calculate the determinant

\[ \begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix} \]

Using the determinant formula, we can expand this determinant:

\[ = 1 \cdot \begin{vmatrix} c & a \\ a & b \end{vmatrix} - 1 \cdot \begin{vmatrix} b & a \\ c & b \end{vmatrix} + 1 \cdot \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

Calculating these 2x2 determinants:

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

### Step 5: Simplify the expression
Combining all terms, we have:

\[ = cb - a^2 - (b^2 - ac) + (ab - c^2) \]

This can be simplified further, but we can also use the fact that the original determinant must equal zero due to the condition for non-trivial solutions.

### Final Result
Since the determinant of the original system is zero, we conclude that:

\[ D = 0 \]