Let for i = 1, 2, 3, $$p_i(x)$$ be a polynomial of degree 2 in $$x$$, $$p'_i(x)$$ and $$p''_i(x)$$ be the first and second order derivatives of $$p_i(x)$$ respectively. Let,
$$A(x) = \begin{bmatrix} p_1(x) & p'_1(x) & p''_1(x) \\ p_2(x) & p'_2(x) & p''_2(x) \\ p_3(x) & p'_3(x) & p''_3(x) \end{bmatrix}$$
and $$B(x) = [A(x)]^T A(x)$$. Then determinant of B(x):

We are given that for $$i = 1, 2, 3$$, $$p_i(x)$$ is a polynomial of degree 2 in $$x$$. The matrix $$A(x)$$ has entries involving $$p_i(x)$$, $$p'_i(x)$$, and $$p''_i(x)$$, and $$B(x) = [A(x)]^T A(x)$$.

Step 1: Write the general form of each polynomial.

Let $$p_i(x) = a_i x^2 + b_i x + c_i$$ where $$a_i \neq 0$$.

Then $$p'_i(x) = 2a_i x + b_i$$ (degree 1)

And $$p''_i(x) = 2a_i$$ (a constant)

Step 2: Write the matrix $$A(x)$$.

$$A(x) = \begin{bmatrix} a_1 x^2 + b_1 x + c_1 & 2a_1 x + b_1 & 2a_1 \\ a_2 x^2 + b_2 x + c_2 & 2a_2 x + b_2 & 2a_2 \\ a_3 x^2 + b_3 x + c_3 & 2a_3 x + b_3 & 2a_3 \end{bmatrix}$$

Step 3: Simplify using column operations.

Apply $$C_1 \to C_1 - x \cdot C_2 + \frac{x^2}{2} \cdot C_3$$ (this does not change the determinant's absolute value — it multiplies det by 1):

The new first column entry for row $$i$$ becomes:

$$(a_i x^2 + b_i x + c_i) - x(2a_i x + b_i) + \frac{x^2}{2}(2a_i)$$

$$= a_i x^2 + b_i x + c_i - 2a_i x^2 - b_i x + a_i x^2 = c_i$$

Next, apply $$C_2 \to C_2 - x \cdot C_3$$:

New second column entry for row $$i$$: $$(2a_i x + b_i) - x(2a_i) = b_i$$

After these operations, the matrix becomes:

$$\begin{bmatrix} c_1 & b_1 & 2a_1 \\ c_2 & b_2 & 2a_2 \\ c_3 & b_3 & 2a_3 \end{bmatrix}$$

This is a constant matrix with no dependence on $$x$$.

Step 4: Conclude about $$\det(A(x))$$.

Since column operations of the type $$C_i \to C_i + \lambda C_j$$ do not change the determinant:

$$\det(A(x)) = \det\begin{bmatrix} c_1 & b_1 & 2a_1 \\ c_2 & b_2 & 2a_2 \\ c_3 & b_3 & 2a_3 \end{bmatrix} = \text{constant}$$

Step 5: Find $$\det(B(x))$$.

Since $$B(x) = A(x)^T A(x)$$:

$$\det(B(x)) = \det(A^T) \cdot \det(A) = [\det(A)]^2 = \text{constant}^2 = \text{constant}$$

Therefore, $$\det(B(x))$$ does not depend on $$x$$.

The correct answer is Option D.