Between the following two statements:

Statement I : Let $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\vec{a} \times \vec{r}=\vec{a} \times \vec{b}$ and $\vec{a} \cdot \vec{r}=0$ is of magnitude $\sqrt{10}$.

Statement II : In a triangle $A B C, \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2}$.

<p>$$\begin{aligned} & \because \quad \forall \text { two vectors } \vec{c} \text { & } \vec{d} \\ & |\vec{c} \times \vec{d}|^2=|\vec{c}|^2|\vec{d}|^2-(\vec{c} \cdot \vec{d})^2 \\ & \text { replacing } \vec{c}=\vec{a} ~\& ~\vec{d}=\vec{r} \\ & \Rightarrow|\vec{a} \times \vec{r}|=|\vec{a}|^2|\vec{r}|^2-(\vec{a} \cdot \vec{r})^2 \\ & \Rightarrow|\vec{a} \times \vec{b}|=|\vec{a}|^2|\vec{r}|^2 \quad(\because \vec{a} \times \vec{r}=\vec{a} \times \vec{b} \text { and } \vec{a} \cdot \vec{r}=0) \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow 35=14|\vec{r}|^2 \\ & \Rightarrow|\vec{r}|=\sqrt{\frac{35}{14}}=\sqrt{\frac{5}{2}} \neq \sqrt{10} \end{aligned}$$</p> <p>$\therefore$ Statement I is incorrect</p> <p>Statement II is correct</p> <p>$\text { (i.e., } \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2} \text { ) }$</p> <p>Proof: $$\because(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \geq 0\quad \text{..... (1)}$$</p> <p>and $$|\overrightarrow{O A}|^2=|\overrightarrow{O B}|^2=|\overrightarrow{O C}|^2=R^2\quad \text{..... (2)}$$</p> <p>Now, using (1), we get</p> <p>$$|\overrightarrow{O A}|^2+|\overrightarrow{O B}|^2+|\overrightarrow{O C}|^2 +2(\overrightarrow{O A} \cdot \overrightarrow{O B}+\overrightarrow{O B} \cdot \overrightarrow{O C}+\overrightarrow{O C} \cdot \overrightarrow{O A}) \geq 0$$</p> <p>$$ \begin{aligned} & \Rightarrow 3 R^2+2 R^2(\cos 2 A+\cos 2 B+\cos 2 C) \geq 0 \\ & \Rightarrow \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2} \end{aligned} $$</p>