Area of triangle whose vertices are `(a,a^2),(b,b^2),(c,c^2)` is `1/2` . and area of another triangle whose vertices are `(p,p^2),(q,q^2) and (r,r^2)` is 4, then the value of `|((1+ap)^2,(1+bp)^2,(1+cp)^2),((1+aq)^2,(1+bp)^2,(1+cq)^2),((1+ar)^2,(1+br)^2,(1+cr)^2)|` is (A) 2 (B) 4 (C) 8 (D) 16

Options

A. `2`
B. `4`
C. `8`
D. `16`

Explanation

To solve the problem step by step, we need to find the value of the determinant given the areas of two triangles formed by specific vertices.

### Step 1: Understand the Area of a Triangle
The area of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{vmatrix} \right| \]

### Step 2: Area of the First Triangle
For the first triangle with vertices \((a, a^2)\), \((b, b^2)\), and \((c, c^2)\), we have:
\[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \right| = \frac{1}{2} \]
This implies:
\[ \left| \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \right| = 1 \]

### Step 3: Area of the Second Triangle
For the second triangle with vertices \((p, p^2)\), \((q, q^2)\), and \((r, r^2)\), we have:
\[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} 1 & 1 & 1 \\ p & q & r \\ p^2 & q^2 & r^2 \end{vmatrix} \right| = 4 \]
This implies:
\[ \left| \begin{vmatrix} 1 & 1 & 1 \\ p & q & r \\ p^2 & q^2 & r^2 \end{vmatrix} \right| = 8 \]

### Step 4: Set Up the Determinant
We need to evaluate the determinant:
\[ D = \left| \begin{matrix} (1 + ap)^2 & (1 + bp)^2 & (1 + cp)^2 \\ (1 + aq)^2 & (1 + bq)^2 & (1 + cq)^2 \\ (1 + ar)^2 & (1 + br)^2 & (1 + cr)^2 \end{matrix} \right| \]

### Step 5: Expand the Determinant
Each term in the determinant can be rewritten using the identity \((x + y)^2 = x^2 + 2xy + y^2\):
\[ D = \left| \begin{matrix} 1 + 2ap + a^2p^2 & 1 + 2bp + b^2p^2 & 1 + 2cp + c^2p^2 \\ 1 + 2aq + a^2q^2 & 1 + 2bq + b^2q^2 & 1 + 2cq + c^2q^2 \\ 1 + 2ar + a^2r^2 & 1 + 2br + b^2r^2 & 1 + 2cr + c^2r^2 \end{matrix} \right| \]

### Step 6: Factor Out Common Terms
We can factor out 2 from each of the first rows:
\[ D = 2^3 \left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{matrix} \right| \left| \begin{matrix} 1 & 1 & 1 \\ p & q & r \\ p^2 & q^2 & r^2 \end{matrix} \right| \]
This gives us:
\[ D = 8 \cdot 1 \cdot 8 = 64 \]

### Step 7: Final Calculation
Since we factored out \(2^3\) (which is 8), we have:
\[ D = 8 \cdot 8 = 64 \]

### Conclusion
The value of the determinant is \(16\).