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 evaluate the determinant given by:

\[ D = \begin{vmatrix} 5 & 4 & 3 \\ x51 & y41 & z31 \\ x & y & z \end{vmatrix} \]

where \(x\), \(y\), and \(z\) are integers in arithmetic progression (AP) between 1 and 9, and \(x51\), \(y41\), and \(z31\) are three-digit numbers.

### Step 1: Understand the Arithmetic Progression
Since \(x\), \(y\), and \(z\) are in AP, we have the relationship:
\[ 2y = x + z \]

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

### Step 3: Substitute into the Determinant
Now we can substitute these expressions into the determinant:
\[ D = \begin{vmatrix} 5 & 4 & 3 \\ 100x + 51 & 100y + 41 & 100z + 31 \\ x & y & z \end{vmatrix} \]

### Step 4: Apply Row Operations
We can simplify the determinant by performing row operations. We will replace the second row by subtracting \(100\) times the third row from it:
\[ R_2 \rightarrow R_2 - 100R_3 \]
This gives us:
\[ D = \begin{vmatrix} 5 & 4 & 3 \\ 51 & 41 & 31 \\ x & y & z \end{vmatrix} \]

### Step 5: Calculate the Determinant
Now we can calculate the determinant:
\[ D = 5 \begin{vmatrix} 41 & 31 \\ y & z \end{vmatrix} - 4 \begin{vmatrix} 51 & 31 \\ x & z \end{vmatrix} + 3 \begin{vmatrix} 51 & 41 \\ x & y \end{vmatrix} \]

### Step 6: Evaluate Each 2x2 Determinant
1. \( \begin{vmatrix} 41 & 31 \\ y & z \end{vmatrix} = 41z - 31y \)

2. \( \begin{vmatrix} 51 & 31 \\ x & z \end{vmatrix} = 51z - 31x \)

3. \( \begin{vmatrix} 51 & 41 \\ x & y \end{vmatrix} = 51y - 41x \)

### Step 7: Substitute Back
Substituting these back into the determinant gives:
\[ D = 5(41z - 31y) - 4(51z - 31x) + 3(51y - 41x) \]

### Step 8: Simplify the Expression
Now we simplify:
\[ D = 205z - 155y - 204z + 124x + 153y - 123x \]
Combining like terms:
\[ D = (205z - 204z) + (-155y + 153y) + (124x - 123x) = z - 2y + x \]

### Step 9: Use the AP Condition
Since \(2y = x + z\), we have:
\[ z - 2y + x = 0 \]

### Conclusion
Thus, the value of the determinant is:
\[ |D| = 0 \]