If a,b,c the three integers lying between 1 and 9 which are in A.P. and a 51, b 41 and c 31 be any three digit numbers , then the value of `|(5,4,3),(a51,b41,c31),(a,b,c)|` is

Options

A. `a+b+c`
B. `a-b-c`
C. 0
D. none

Answer

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

\[ \left| \begin{array}{ccc} 5 & 4 & 3 \\ a51 & b41 & c31 \\ a & b & c \end{array} \right| \]

where \( a, b, c \) are integers between 1 and 9 that are in arithmetic progression (A.P.).

### Step 1: Express the three-digit numbers
The three-digit numbers can be expressed as:
- \( a51 = 100a + 50 + 1 = 100a + 51 \)
- \( b41 = 100b + 40 + 1 = 100b + 41 \)
- \( c31 = 100c + 30 + 1 = 100c + 31 \)

### Step 2: Rewrite the determinant
Now, substituting these expressions into the determinant, we have:

\[ \left| \begin{array}{ccc} 5 & 4 & 3 \\ 100a + 51 & 100b + 41 & 100c + 31 \\ a & b & c \end{array} \right| \]

### Step 3: Simplify the determinant
We can perform row operations to simplify the determinant. Let's subtract the third row from the second row:

\[ \left| \begin{array}{ccc} 5 & 4 & 3 \\ (100a + 51 - a) & (100b + 41 - b) & (100c + 31 - c) \\ a & b & c \end{array} \right| \]

This simplifies to:

\[ \left| \begin{array}{ccc} 5 & 4 & 3 \\ 99a + 51 & 99b + 41 & 99c + 31 \\ a & b & c \end{array} \right| \]

### Step 4: Factor out common terms
Now, we can factor out 99 from the second row:

\[ \left| \begin{array}{ccc} 5 & 4 & 3 \\ 99(a + \frac{51}{99}) & 99(b + \frac{41}{99}) & 99(c + \frac{31}{99}) \\ a & b & c \end{array} \right| \]

This gives us:

\[ 99 \left| \begin{array}{ccc} 5 & 4 & 3 \\ a + \frac{51}{99} & b + \frac{41}{99} & c + \frac{31}{99} \\ a & b & c \end{array} \right| \]

### Step 5: Use properties of determinants
Now, we can perform another row operation by subtracting the first row multiplied by a suitable factor from the second row to create zeros.

We can also use the property that if \( a, b, c \) are in A.P., then \( 2b = a + c \).

### Step 6: Calculate the determinant
After performing the necessary row operations and simplifying, we will find that:

\[ \text{det} = a + c - 2b = 0 \]

### Conclusion
Since \( a, b, c \) are in A.P., we have \( a + c - 2b = 0 \). Therefore, the value of the determinant is:

\[ \boxed{0} \]