$$\begin{aligned} & \because \quad \forall \text { two vectors } \vec{c} \text { & } \vec{d} \\ & |\vec{c} \times \vec{d}|^2=|\vec{c}|^2|\vec{d}|^2-(\vec{c} \cdot \vec{d})^2 \\ & \text { replacing } \vec{c}=\vec{a} ~\& ~\vec{d}=\vec{r} \\ & \Rightarrow|\vec{a} \times \vec{r}|=|\vec{a}|^2|\vec{r}|^2-(\vec{a} \cdot \vec{r})^2 \\ & \Rightarrow|\vec{a} \times \vec{b}|=|\vec{a}|^2|\vec{r}|^2 \quad(\because \vec{a} \times \vec{r}=\vec{a} \times \vec{b} \text { and } \vec{a} \cdot \vec{r}=0) \end{aligned}$$

$$\begin{aligned} & \Rightarrow 35=14|\vec{r}|^2 \\ & \Rightarrow|\vec{r}|=\sqrt{\frac{35}{14}}=\sqrt{\frac{5}{2}} \neq \sqrt{10} \end{aligned}$$

$$\therefore$$ Statement I is incorrect

Statement II is correct

$$\text { (i.e., } \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2} \text { ) }$$

Proof: $$\because(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \geq 0\quad \text{..... (1)}$$

and $$|\overrightarrow{O A}|^2=|\overrightarrow{O B}|^2=|\overrightarrow{O C}|^2=R^2\quad \text{..... (2)}$$

Now, using (1), we get

$$|\overrightarrow{O A}|^2+|\overrightarrow{O B}|^2+|\overrightarrow{O C}|^2 +2(\overrightarrow{O A} \cdot \overrightarrow{O B}+\overrightarrow{O B} \cdot \overrightarrow{O C}+\overrightarrow{O C} \cdot \overrightarrow{O A}) \geq 0$$

$$ \begin{aligned} & \Rightarrow 3 R^2+2 R^2(\cos 2 A+\cos 2 B+\cos 2 C) \geq 0 \\ & \Rightarrow \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2} \end{aligned} $$