If `|{:(a-b,b-c,c-a),(x-y,y-z,z-x),(p-q,q-r,r-p):}|=m|{:(c,a,b),(z,x,y),(r,p,q):}|," then m"=`

Options

A. `2`
B. `1`
C. `0`
D. None of these

Answer

To solve the given problem, we need to evaluate the determinants and find the value of \( m \) such that:

\[ |{:(a-b, b-c, c-a), (x-y, y-z, z-x), (p-q, q-r, r-p):}| = m |{:(c, a, b), (z, x, y), (r, p, q):}| \]

### Step-by-Step Solution:

1. **Define the Determinants:**
Let \( D_1 = |{:(a-b, b-c, c-a), (x-y, y-z, z-x), (p-q, q-r, r-p):}| \) and \( D_2 = |{:(c, a, b), (z, x, y), (r, p, q):}| \).

2. **Evaluate \( D_1 \):**
We will simplify \( D_1 \) using properties of determinants. We can perform column operations to simplify the determinant.

- Add the second and third columns to the first column:
\[ C_1 \to C_1 + C_2 + C_3 \]
- This results in:
\[ D_1 = |{:(a-b + (b-c) + (c-a), b-c, c-a), (x-y + (y-z) + (z-x), y-z, z-x), (p-q + (q-r) + (r-p), q-r, r-p):}| \]
- The first column simplifies to:
\[ 0, 0, 0 \]
- Therefore, we have:
\[ D_1 = |{:(0, b-c, c-a), (0, y-z, z-x), (0, q-r, r-p):}| \]

3. **Expand \( D_1 \):**
Since the first column consists entirely of zeros, the determinant \( D_1 \) evaluates to:
\[ D_1 = 0 \]

4. **Evaluate \( D_2 \):**
The determinant \( D_2 \) is a standard determinant and does not need simplification for this problem since we only need to compare it with \( D_1 \).

5. **Set Up the Equation:**
From the initial equation, we have:
\[ D_1 = m \cdot D_2 \]
Substituting the values we found:
\[ 0 = m \cdot D_2 \]

6. **Solve for \( m \):**
Since \( D_2 \) is not zero (assuming \( c, a, b, z, x, y, r, p, q \) are distinct), the only solution for \( m \) that satisfies the equation is:
\[ m = 0 \]

### Conclusion:
Thus, the value of \( m \) is \( 0 \).