If x,y and z are the integers in AP lying between 1 and 9 and x 51, y 41 and z 31 are three digits number the value of `|{:(5,4,3),(x51,y41,z31),(x,y,z):}|` is

Options

A. x+y+z
B. x-y+z
C. 0
D. None of these

Answer

To solve the problem, we need to find the value of the determinant \(|{:(5,4,3),(x51,y41,z31),(x,y,z):}|\) where \(x\), \(y\), and \(z\) are integers in arithmetic progression (AP) between 1 and 9.

### Step-by-Step Solution:

1. **Understanding the Arithmetic Progression**:
Since \(x\), \(y\), and \(z\) are in AP, we have:
\[ y = \frac{x + z}{2} \]
This implies:
\[ 2y = x + z \]

2. **Expressing the Three-Digit Numbers**:
The three-digit numbers can be expressed as:
- \(x51 = 100x + 50 + 1 = 100x + 51\)
- \(y41 = 100y + 40 + 1 = 100y + 41\)
- \(z31 = 100z + 30 + 1 = 100z + 31\)

3. **Setting Up the Determinant**:
The determinant can be written as:
\[ D = \begin{vmatrix} 5 & 4 & 3 \\ 100x + 51 & 100y + 41 & 100z + 31 \\ x & y & z \end{vmatrix} \]

4. **Row Operations**:
We can simplify the determinant using row operations. We will perform the operation:
\[ R_2 \rightarrow R_2 - 100R_3 - 10R_1 \]
This gives us:
- First row remains \( (5, 4, 3) \)
- Second row becomes:
\[ (100x + 51 - 100x - 50, 100y + 41 - 100y - 40, 100z + 31 - 100z - 30) = (1, 1, 1) \]
- Third row remains \( (x, y, z) \)

Thus, the determinant simplifies to:
\[ D = \begin{vmatrix} 5 & 4 & 3 \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} \]

5. **Calculating the Determinant**:
Now we can compute the determinant:
\[ D = 5 \begin{vmatrix} 1 & 1 \\ y & z \end{vmatrix} - 4 \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} + 3 \begin{vmatrix} 1 & 1 \\ x & y \end{vmatrix} \]

Calculating each of these 2x2 determinants:
- \( \begin{vmatrix} 1 & 1 \\ y & z \end{vmatrix} = z - y \)
- \( \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} = z - x \)
- \( \begin{vmatrix} 1 & 1 \\ x & y \end{vmatrix} = y - x \)

Therefore,
\[ D = 5(z - y) - 4(z - x) + 3(y - x) \]
Simplifying this:
\[ D = 5z - 5y - 4z + 4x + 3y - 3x = z - 2y + x \]

6. **Substituting the AP Condition**:
From the AP condition, we know \(x + z = 2y\). Thus:
\[ D = (x + z) - 2y = 2y - 2y = 0 \]

### Final Answer:
The value of the determinant is \(0\).