The area of a $\vartriangle ABC$ with given vertices $(a,a^2),(b,b^2),(c,c^2)$ is $\frac{1}{4}$ $sq. units$ and area of another $\vartriangle PQR$ with given vertices $(p,p^2),(q,q^2),(r,r^2)$ is $3$ $sq. units$.
Then what is the value of $$ \begin{vmatrix} (1+ap)^2 & (1+bp)^2 & (1+cp)^2 \\ (1+aq)^2 & (1+bq)^2 & (1+cq)^2 \\ (1+ar)^2 & (1+br)^2 & (1+cr)^2 \\ \end{vmatrix} $$
I could not even begin attempting it , i don't know where to begin from,someone kindly help.
Let $A, B, C$, $P, Q, R$ be the $6$ column vectors
$$ \begin{cases} A^T = (1, \sqrt{2}a, a^2),\\ B^T = (1, \sqrt{2}b, b^2),\\ C^T = (1, \sqrt{2}c, c^2) \end{cases} \quad\text{ and }\quad \begin{cases} P^T = (1, \sqrt{2}p, p^2),\\ Q^T = (1, \sqrt{2}q, q^2),\\ R^T = (1, \sqrt{2}r, r^2) \end{cases} $$ Using identites of the form $$(1+ap)^2 = 1 + 2ap + a^2p^2 = 1\cdot 1 + \sqrt{2}a\cdot\sqrt{2}p + a^2\cdot p^2 = A\cdot P$$ We can rewrite the determinant at hand as
$$\Delta \stackrel{def}{=}\begin{vmatrix} (1+ap)^2 & (1+bp)^2 & (1+cp)^2 \\ (1+aq)^2 & (1+bq)^2 & (1+cq)^2 \\ (1+ar)^2 & (1+br)^2 & (1+cr)^2 \\ \end{vmatrix} = \begin{vmatrix} A\cdot P & B\cdot P & C\cdot P \\ A\cdot Q & B\cdot Q & C\cdot Q \\ A\cdot R & B \cdot R & C\cdot R \\ \end{vmatrix} $$ Notice the matrix for rightmost determinant is a product of two $3 \times 3$ matrices $$ \begin{bmatrix} A\cdot P & B\cdot P & C\cdot P \\ A\cdot Q & B\cdot Q & C\cdot Q \\ A\cdot R & B \cdot R & C\cdot R \\ \end{bmatrix} = \left[ P, Q, R\right]^T \left[A, B, C\right] $$ This leads to (up to a sign), $$\Delta = \begin{vmatrix} 1 & \sqrt{2}p & p^2 \\ 1 & \sqrt{2}q & q^2 \\ 1 & \sqrt{2}r & r^2 \\ \end{vmatrix} \begin{vmatrix} 1 & 1 & 1\\ \sqrt{2}a & \sqrt{2}b & \sqrt{2}c \\ a^2 & b^2 & c^2 \\ \end{vmatrix} = 2 \begin{vmatrix} 1 & p & p^2 \\ 1 & q & q^2 \\ 1 & r & r^2 \\ \end{vmatrix} \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix} = 2(2\times 3)(2\times\frac14) = 6 $$
The oriented area of a triangle with vertices
$$A(x_A,y_A), B(x_B,y_B), C(x_C,y_C)$$
is known to be given by the following formula: $$\tag{1}Area(ABC)=\tfrac12 \begin{vmatrix}x_A & x_B & x_C\\y_A & y_B & y_C\\1 & 1 & 1\end{vmatrix}$$
(this area is positive if the orientation of ABC is direct, negative otherwise).
See here for a proof of (1).
In fact, you are working on non oriented areas in this issue ;
therefore, the following calculations are done with absolute values.
Thus, taking (1) into account, the hypotheses can be written:
$$\tag{2}\begin{vmatrix}a & b & c\\a^2 & b^2 & c^2\\1 & 1 & 1\end{vmatrix}=\dfrac12 \ \ \ \ \ \text{and} \ \ \ \ \ \begin{vmatrix}p & q & r\\p^2 & q^2 & r^2\\1 & 1 & 1\end{vmatrix}=6.$$
Consider the following matrix decomposition:
$$\underbrace{\begin{pmatrix} (1+ap)^2 & (1+bp)^2 & (1+cp)^2 \\ (1+aq)^2 & (1+bq)^2 & (1+cq)^2 \\ (1+ar)^2 & (1+br)^2 & (1+cr)^2 \\ \end{pmatrix}}_C=\underbrace{\begin{pmatrix}1 & 2a & a^2\\1 & 2b & b^2\\1 & 2c & c^2\end{pmatrix}}_A\underbrace{\begin{pmatrix}1 & 1 & 1\\p & q & r\\p^2 & q^2 & r^2\end{pmatrix}}_B$$
Using some well known properties of determinants, we obtain, using (2) :
$$|det(C)|=|det(A)||det(B)| \ = \ 2 (\tfrac12) \times 6 \ = \ 6.$$
Remark:
1) Matrices like $B$ are called Vandermonde matrices. Their determinant is found in many applications; see for example slides 22 and following of (http://www.math.tamu.edu/~yvorobet/MATH304-2011A/Lect1-07web.pdf).
2) The way I have found decomposition (2): I expanded at first the determinant and I saw there was an intermixing of the different combinations of all the occurences of $a,b,c$ with $p,q,r$, something that occurs in the product of matrices.
