If `-a^2x+aby+acz=0,abx-b^2y+bcz=0 and acx+bcy-c^2z=0` have non-trivial solution then find the value of `|{:(0,c,b),(c,0,a),(b,a,0):}|`

Answer

To solve the problem, we need to find the value of the determinant given by the matrix:

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

### Step 1: Write down the determinant

The determinant can be expressed as:

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

### Step 2: Expand the determinant

We can use the cofactor expansion along the first row:

\[ D = 0 \cdot \begin{vmatrix} 0 & a \\ a & 0 \end{vmatrix} - c \cdot \begin{vmatrix} c & a \\ b & 0 \end{vmatrix} + b \cdot \begin{vmatrix} c & 0 \\ b & a \end{vmatrix} \]

The first term is zero, so we only need to calculate the remaining two terms:

\[ D = -c \cdot \begin{vmatrix} c & a \\ b & 0 \end{vmatrix} + b \cdot \begin{vmatrix} c & 0 \\ b & a \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants

1. For the first determinant:

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

2. For the second determinant:

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

### Step 4: Substitute back into the determinant expression

Now substituting these results back into our expression for \(D\):

\[ D = -c(-ab) + b(ac) = cab + abc = 2abc \]

### Step 5: Set the determinant to zero

Since we are looking for the conditions under which the original system has a non-trivial solution, we set the determinant \(D\) equal to zero:

\[ 2abc = 0 \]

### Step 6: Conclusion

This implies that at least one of \(a\), \(b\), or \(c\) must be zero. Therefore, the value of the determinant \(D\) is:

\[ \boxed{0} \]