If x+y+z=0=a+b+c, then `|{:(xa,yb,zc),(yc,za,xb),(zb,xc,ya):}|=`

Options

A. 0
B. `xa+yb+zc`
C. 1
D. None of these

Answer

To solve the determinant given the conditions \(x + y + z = 0\) and \(a + b + c = 0\), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant given by:
\[ D = \begin{vmatrix} xa & yb & zc \\ yc & za & xb \\ zb & xc & ya \end{vmatrix} \]

### Step 2: Apply the Determinant Properties
We can expand the determinant using the first row. The determinant can be calculated as:
\[ D = xa \begin{vmatrix} za & xb \\ xc & ya \end{vmatrix} - yb \begin{vmatrix} yc & xb \\ zb & ya \end{vmatrix} + zc \begin{vmatrix} yc & za \\ zb & xc \end{vmatrix} \]

### Step 3: Calculate Each 2x2 Determinant
Calculating the first 2x2 determinant:
\[ \begin{vmatrix} za & xb \\ xc & ya \end{vmatrix} = (za)(ya) - (xb)(xc) = zay - bxy \]

Calculating the second 2x2 determinant:
\[ \begin{vmatrix} yc & xb \\ zb & ya \end{vmatrix} = (yc)(ya) - (xb)(zb) = yac - bzx \]

Calculating the third 2x2 determinant:
\[ \begin{vmatrix} yc & za \\ zb & xc \end{vmatrix} = (yc)(xc) - (za)(zb) = ycx - zab \]

### Step 4: Substitute Back into the Determinant
Now substituting these back into the expression for \(D\):
\[ D = xa(zay - bxy) - yb(yac - bzx) + zc(ycx - zab) \]

### Step 5: Expand and Simplify
Expanding the terms:
\[ D = xazay - xabxy - ybyac + ybbzx + zcycx - zcab \]

### Step 6: Factor Out Common Terms
Notice that we can factor out \(xyz\) from the terms:
\[ D = xyz(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step 7: Use the Given Conditions
Since \(a + b + c = 0\), we can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
This means:
\[ a^3 + b^3 + c^3 = 3abc \]

### Step 8: Conclude the Result
Thus, we have:
\[ D = xyz \cdot 0 = 0 \]

### Final Answer
Therefore, the value of the determinant is:
\[ \boxed{0} \]