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 $$ABC$$, $$\cos 2A + \cos 2B + \cos 2C \ge -\frac{3}{2}$$.

Statement I: $$\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$$, $$\vec{b}=2\hat{i}+\hat{j}-\hat{k}$$. $$\vec{a}\times\vec{r}=\vec{a}\times\vec{b}$$ means $$\vec{a}\times(\vec{r}-\vec{b})=\vec{0}$$, so $$\vec{r}-\vec{b}=\lambda\vec{a}$$.

$$\vec{r}=\vec{b}+\lambda\vec{a}=(2+\lambda)\hat{i}+(1+2\lambda)\hat{j}+(-1-3\lambda)\hat{k}$$.

$$\vec{a}\cdot\vec{r}=0$$: $$(2+\lambda)+2(1+2\lambda)-3(-1-3\lambda)=0$$. $$2+\lambda+2+4\lambda+3+9\lambda=0$$. $$14\lambda+7=0$$, $$\lambda=-1/2$$.

$$\vec{r}=3/2\hat{i}+0\hat{j}+1/2\hat{k}$$. $$|\vec{r}|=\sqrt{9/4+1/4}=\sqrt{10/4}=\sqrt{10}/2$$.

$$|\vec{r}| = \sqrt{10}/2 \neq \sqrt{10}$$. Statement I is incorrect.

Statement II: $$\cos 2A+\cos 2B+\cos 2C \geq -3/2$$.

$$\cos 2A+\cos 2B+\cos 2C = -1-4\cos A\cos B\cos C$$. Min of $$-1-4\cos A\cos B\cos C$$: max of $$\cos A\cos B\cos C = 1/8$$ (at equilateral). Min = $$-1-4(1/8) = -3/2$$. So $$\geq -3/2$$. Statement II is correct.

The correct answer is Option 1.