Suppose a,b,c are in A.P if p,q,r are also in A.P., then value of
`Delta=|(x^(2)+a,x+p,c),(x^(2)+b,x+q,b),(x^(2)+c,x+r,a)|`
is dependent on

Options

A. x
B. a,b,c are in H.P
C. p,q,r
D. No

Answer

To solve the problem, we need to evaluate the determinant \( \Delta \) given by:

\[ \Delta = \begin{vmatrix} x^2 + a & x + p & c \\ x^2 + b & x + q & b \\ x^2 + c & x + r & a \end{vmatrix} \]

### Step 1: Identify the properties of A.P.
Since \( a, b, c \) are in arithmetic progression (A.P.), we have:
\[ b - a = c - b \implies 2b = a + c \]

Similarly, since \( p, q, r \) are also in A.P., we have:
\[ q - p = r - q \implies 2q = p + r \]

### Step 2: Perform row operations
We can simplify the determinant by performing row operations. Specifically, we can subtract the first row from the second and third rows:

\[ R_2 \rightarrow R_2 - R_1 \]
\[ R_3 \rightarrow R_3 - R_1 \]

This gives us:

\[ \Delta = \begin{vmatrix} x^2 + a & x + p & c \\ b - a & q - p & b - c \\ c - a & r - p & a - c \end{vmatrix} \]

### Step 3: Simplify the determinant
Now, we can express the second and third rows in terms of their differences:

1. The second row becomes:
\[ (b - a, q - p, b - c) \]
2. The third row becomes:
\[ (c - a, r - p, a - c) \]

### Step 4: Evaluate the determinant
The determinant can be evaluated using the properties of determinants. If two rows or two columns are proportional or equal, the determinant will be zero.

Notice that if we factor out common terms from the rows, we can find that the determinant simplifies to zero due to linear dependence.

### Conclusion
Thus, we conclude that:

\[ \Delta = 0 \]

This means that the value of \( \Delta \) does not depend on \( x, a, b, c, p, q, r \).

### Final Answer
The value of \( \Delta \) is dependent on nothing, as it evaluates to zero.

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