I'm going to write a lower-size version of my question. The Solution or hint for this one might be sufficient as well.
Let $G=\begin{bmatrix} A & B\\ C&D \end{bmatrix} $ where $$A_{11}=2C_{1}^{\frac{1}{2}}(\langle\eta_1,\eta_1\rangle)=2$$ $$A_{12}=2C_{1}^{\frac{1}{2}}(\langle\eta_1,\eta_2\rangle) +i(\eta_{11}\eta_{22}-\eta_{12}\eta_{21})C_{0}^{\frac{3}{2}}(\langle\eta_1,\eta_2\rangle)$$ $$A_{21}=\overline{A_{12}}$$ $$A_{22}=2C_{1}^{\frac{1}{2}}(\langle\eta_2,\eta_2\rangle)=2$$
So $A$ is a hermitian matrix.
$$B_{11}=0$$ $$B_{12}=(\eta_{11}\eta_{23}-\eta_{13}\eta_{21})C_{0}^{\frac{3}{2}}(\langle\eta_1,\eta_2\rangle) +i(\eta_{12}\eta_{23}-\eta_{13}\eta_{22})C_{0}^{\frac{3}{2}}(\langle\eta_1,\eta_2\rangle)$$
$$B_{21}=-B_{12}$$ $$B_{22}=0$$
$$C=-\overline{B}$$ i.e., $C_{11}=C_{22}=0, $ $C_{12}=-\overline{B_{12}}, $ and $C_{21}=\overline{B_{12}}$, and $$D=\overline{A}$$ i.e., $D_{11}=D_{22}=2, $ $D_{12}=\overline{A_{12}}, $ and $D_{21}=A_{12}$.
where $\eta_{i}=(\eta_{i1},\eta_{i2},\eta_{i3})$ (we have only two points) are the points on the unit sphere. And $\langle\cdot,\cdot\rangle$ is the inner product.
Prove that $G$ is positive definite for almost every random choice of points.
Edit after a comment
\begin{align*} \det(G)&=16+\vert A_{12}\vert^{4}-\vert B_{12}\vert^{4}-8(\vert A_{12}\vert^{2}+\vert B_{12}\vert^{2})\\ &\;\;\; +2\vert A_{12}\vert^{2}\vert B_{12}\vert^{2}+(\vert A_{12}\vert^{2}S+\vert B_{12}\vert^{2}T)i-TS \end{align*} where \begin{align*} \vert A_{12}\vert^{2}&=4 \big(C_{1}^{\frac{1}{2}}(\langle \eta_{1},\eta_{2} \rangle)\big)^{2}+(\eta_{11}\eta_{22}-\eta_{12}\eta_{21})^{2} \big( C_{0}^{\frac{3}{2}}(\langle \eta_{1},\eta_{2} \rangle)\big)^{2}\\ \vert B_{12}\vert^{2}&=(\eta_{11}\eta_{23}-\eta_{13}\eta_{21})^{2} \big( C_{0}^{\frac{3}{2}}(\langle\eta_1,\eta_2\rangle)\big)^{2} +(\eta_{12}\eta_{23}-\eta_{13}\eta_{22})^{2} \big( C_{0}^{\frac{3}{2}}(\langle\eta_1,\eta_2\rangle)\big)^{2}\\ T&=4\, C_{1}^{\frac{1}{2}}(\langle \eta_{1},\eta_{2} \rangle)\, C_{0}^{\frac{3}{2}}(\langle \eta_{1},\eta_{2} \rangle)\, (\eta_{11}\eta_{22}-\eta_{12}\eta_{21})\\ S&=2\,\big(C_{0}^{\frac{3}{2}}(\langle \eta_{1},\eta_{2} \rangle)\big)^{2}\, (\eta_{11}\eta_{23}-\eta_{13}\eta_{21})\, (\eta_{12}\eta_{23}-\eta_{13}\eta_{22})\\ \end{align*}
I understand that $C_{1}^{\frac{1}{2}}(\langle \eta_{1},\eta_{2} \rangle)=\langle \eta_{1},\eta_{2} \rangle=\cos(\theta)$ and $C_{0}^{\frac{3}{2}}(\langle \eta_{1},\eta_{2} \rangle)=1$. But even after this it's not obvious that the determinant is nonzero.
In this particular case, I think the answer is not very hard.
First, your Gegenbauer polynomials are very simple $$ C_1^{\frac{1}{2}}(x)=x,\quad C_0^{\frac{3}{2}}(x)=1. $$
Second, without loss of generality, you can always make one point $\eta_1$ on $S^2$ to be the north pole, and write the other one in spherical coordinates. Let me use $\alpha$ and $\beta$ for the usual $\phi$ and $\theta$ angles. (See the end for a further simplification.) Then clearly $$ \langle \eta_1, \eta_2\rangle =\cos\alpha. $$ Putting them together as rows, we get $$ \begin{pmatrix} 0 & 0 & 1\\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{pmatrix}, $$ then the entries in your matrix are nothing but certain $2\times 2$ minors (they are also coordinates of the cross product $\eta_1\times\eta_2$).
Then you can work out your matrix $A$ and $B$ concretely. \begin{gather} A = \begin{pmatrix} 2 & 2\cos\alpha\\ 2\cos\alpha & 2 \end{pmatrix},\\ B=\begin{pmatrix} 0 & -(\sin\alpha) (\cos\beta + i\sin\beta)\\ (\sin\alpha) (\cos\beta + i\sin\beta) & 0 \end{pmatrix} =\begin{pmatrix} 0 & -(\sin\alpha) e^{i\beta}\\ (\sin\alpha) e^{i\beta} & 0 \end{pmatrix}. \end{gather}
Here $A=\bar A^T$ is Hermitian, and $B=-B^T$ is skew-symmetric.
Your total matrix $$G=\begin{pmatrix} A & B\\ -\bar B & \bar A \end{pmatrix} =\begin{pmatrix} A & B\\ \bar B^T & \bar A \end{pmatrix} $$ is a Hermitian matrix. To check whether it is positive-definite, we only need to check if it has all positive principal minors.
Most concretely, we have $$ G = \begin{pmatrix} 2 & 2\cos\alpha & 0 & -(\sin\alpha) e^{i\beta}\\ 2\cos\alpha & 2 & (\sin\alpha) e^{i\beta} & 0\\ 0 & (\sin\alpha) e^{-i\beta} & 2 & 2\cos\alpha\\ - (\sin\alpha) e^{-i\beta} & 0 & 2\cos\alpha & 2 \end{pmatrix}. $$ Let $g_i$ be the $i$th principal minor of $G$, that is, the determinant of the submatrix of $G$ from the first $i$ rows and columns. Then a direct calculation gives \begin{align*} g_1 &= 2,\\ g_2 &= 4\sin^2\alpha,\\ g_3 &= 6\sin^2\alpha,\\ g_4 &= 9\sin^4\alpha. \end{align*} Therefore the matrix $G$ is positive-definite when $\sin\alpha\neq 0$. That happens when $\eta_2$ is not the north or south pole.
Therefore, the final conclusion is that your $G$ is positive-definite whenever $\eta_1\neq \pm \eta_2$ on $S^2$.
Added for simplification. We can actually just let $\beta=0$ in the above calculation. (This corresponds to choosing the coordinate system so that $\eta_1$ is at the north pole, and $\eta_2$ is in the $xz-$plane.) Then $e^{i\beta}=1$, and the matrix $G$ is totally real and symmetric as $$ G=\begin{pmatrix} A & B\\ -B & A \end{pmatrix}. $$ Such a matrix is actually the corresponding real form of the Hermitian matrix $$ H= A-iB = \begin{pmatrix} 2 & 2\cos\alpha + i\sin\alpha\\ 2\cos\alpha - i\sin\alpha & 2 \end{pmatrix}. $$ Here $G$ is positive-definite as a symmetric real matrix iff $H$ is positive-definite as a Hermitian matrix. The reason is that for $$\vec w=(w_1, w_2)^T = (x_1+iy_1, x_2+iy_2)^T,$$ let $$ \vec v = (x_1, x_2, y_1, y_2)^T, $$ and one can check that $$ \vec v^T G \vec v = \vec w^T H \vec w.$$ Now the pricipal minors of $H$ are $$ h_1 = 2,\quad h_2 = 3\sin^2\alpha. $$ We arrive at the same conclusion as before, although this time the determinant calculation is very easy.