If a,b,c are in A.P. and p is a real number and
`Delta=|(p+c,p+2,p+a),(p+b,p+5,p+b),(p+a,p+8,p+c)|`
then `Delta` equals

Options

A. `-p^(3)`
B. `p^(3)`
C. `p^(3)-2abc`
D. 0

Answer

To solve the determinant problem where \( \Delta = \begin{vmatrix} p+c & p+2 & p+a \\ p+b & p+5 & p+b \\ p+a & p+8 & p+c \end{vmatrix} \) and \( a, b, c \) are in Arithmetic Progression (A.P.), we will follow these steps:

### Step 1: Understand the Condition of A.P.
Since \( a, b, c \) are in A.P., we have the relation:
\[ b = \frac{a+c}{2} \]

### Step 2: Substitute the A.P. Relation into the Determinant
We can rewrite the determinant using the relation \( b = \frac{a+c}{2} \):
\[ \Delta = \begin{vmatrix} p+c & p+2 & p+a \\ \frac{a+c}{2} + p & p+5 & \frac{a+c}{2} + p \\ p+a & p+8 & p+c \end{vmatrix} \]

### Step 3: Simplify the Determinant
Now, we can simplify the determinant by performing row operations. We can subtract the first row from the second row:
\[ \Delta = \begin{vmatrix} p+c & p+2 & p+a \\ \frac{a+c}{2} - (c) & 3 & \frac{a+c}{2} - (a) \\ p+a & p+8 & p+c \end{vmatrix} \]

This gives us:
\[ \Delta = \begin{vmatrix} p+c & p+2 & p+a \\ \frac{a-c}{2} & 3 & \frac{c-a}{2} \\ p+a & p+8 & p+c \end{vmatrix} \]

### Step 4: Check for Row or Column Similarities
Notice that if we manipulate the columns or rows further, we can check if any two rows or columns are identical. If they are, the determinant will equal zero.

### Step 5: Apply the Determinant Property
If we find that any two rows or columns are identical, we can conclude:
\[ \Delta = 0 \]

### Conclusion
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = 0 \]