Let $$\mathbb{R}^2$$ denote $$\mathbb{R} \times \mathbb{R}$$. Let $$S = \{(a, b, c) : a, b, c \in \mathbb{R}$$ and $$ax^2 + 2bxy + cy^2 \gt 0$$ for all $$(x, y) \in \mathbb{R}^2 - \{(0, 0)\}\}$$. Then which of the following statements is (are) TRUE?

A quadratic form $$Q(x,y)=ax^2+2bxy+cy^2$$ is positive for every non-zero $$(x,y)\in\mathbb{R}^2$$ (that is, positive definite) iff its associated symmetric matrix $$\begin{pmatrix}a & b\\[2pt] b & c\end{pmatrix}$$ is positive definite.

Sylvester’s criterion for a $$2\times2$$ symmetric matrix gives the necessary and sufficient conditions
    $$a\gt0,\qquad c\gt0,\qquad ac-b^2\gt0\;$$ $$-(1)$$
Hence $$S=\{(a,b,c):a\gt0,\;c\gt0,\;ac-b^2\gt0\}.$$

We check each option using $$-(1)$$.

Option A
Take $$(a,b,c)=\left(2,\frac72,6\right).$$
    $$ac-b^2=2\cdot6-\left(\frac72\right)^2=12-\frac{49}{4}=\frac{48-49}{4}=-\frac14\lt0.$$
Condition $$ac-b^2\gt0$$ fails, so $$\left(2,\frac72,6\right)\notin S.$$
Option A is false.

Option B
Assume $$\left(3,b,\frac1{12}\right)\in S.$$ Applying $$-(1)$$:
    $$3\gt0,\quad \frac1{12}\gt0\quad(\text{true}),$$
    $$3\cdot\frac1{12}-b^2\gt0\;\Longrightarrow\; \frac14-b^2\gt0\;\Longrightarrow\;b^2\lt\frac14.$$
Thus $$|b|\lt\frac12,$$ that is, $$|2b|\lt1.$$ Option B is true.

Option C
For any $$(a,b,c)\in S$$ the coefficient matrix of the given linear system
    $$\begin{cases}ax+by=1\\ bx+cy=-1\end{cases}$$
is $$\begin{pmatrix}a&b\\ b&c\end{pmatrix}.$$
From $$-(1)$$ its determinant $$ac-b^2$$ is strictly positive, hence non-zero. Therefore the matrix is invertible and the system has a unique solution for every $$(a,b,c)\in S.$$ Option C is true.

Option D
For the system
    $$\begin{cases}(a+1)x+by=0\\ bx+(c+1)y=0\end{cases}$$
the coefficient matrix is $$\begin{pmatrix}a+1&b\\ b&c+1\end{pmatrix}.$$ Its determinant is
    $$(a+1)(c+1)-b^2=ac-b^2 + a+c+1.$$
Using $$-(1)$$, $$ac-b^2\gt0$$ and $$a,c\gt0$$, hence
    $$(a+1)(c+1)-b^2\gt0.$$
The determinant is always positive, so the matrix is invertible and the system possesses a unique solution for every $$(a,b,c)\in S.$$ Option D is true.

Therefore the correct statements are:
Option B (|2b| < 1), Option C (unique solution), and Option D (unique solution).

Answer: Option B, Option C, Option D.