Let $S$ be a circle with its center at point O and a radius of 1. Let $\triangle \mathrm { ABC }$ be a triangle such that all its vertices are on $S$ and $\mathrm { AB } : \mathrm { AC } = 3 : 2$. As shown in the figure, let D be a point on the extension of side BC and $k$ be the number where
$$\mathrm { BC } : \mathrm { CD } = 2 : k .$$
Moreover, set
$$\overrightarrow { \mathrm { OA } } = \vec { a } , \quad \overrightarrow { \mathrm { OB } } = \vec { b } , \quad \overrightarrow { \mathrm { OC } } = \vec { c }$$
Answer the following questions.
(1) When we express $\overrightarrow { \mathrm { OD } }$ in terms of $\vec { b } , \vec { c }$ and $k$, we have
$$\overrightarrow { \mathrm { OD } } = \left( \frac { k } { \mathbf { A } } + \mathbf { B } \right) \vec { c } - \frac { k } { \mathbf { C } } \vec { b }$$
(2) Since the equality
$$| \vec { b } - \vec { a } | = \frac { \mathbf { D } } { \mathbf { E } } | \vec { c } - \vec { a } |$$
holds, by expressing the inner product $\vec { a } \cdot \vec { b }$ in terms of the inner product $\vec { a } \cdot \vec { c }$, we have
$$\vec { a } \cdot \vec { b } = \frac { \mathbf { F } } { \mathbf { G } } \vec { a } \cdot \vec { c } - \frac { \mathbf { H } } { \mathbf { I } }$$
(3) It follows that when the tangent to $S$ at the point A passes through the point D,
$$k = \frac { \mathbf { J } } { \mathbf { K } } .$$