Let $$\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$$ and $$\vec{b} = \hat{j} + 2\hat{k}$$. If $$\vec{r}$$ is a vector such that $$\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$$ and $$\vec{r} \cdot \vec{a} = 0$$, then $$|3\vec{r}|^2$$ is equal to :

$$\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$$

$$\vec{r} \times \vec{a} - \vec{b} \times \vec{a} = \vec{0}$$

$$(\vec{r} - \vec{b}) \times \vec{a} = \vec{0}$$

$$\vec{r} - \vec{b} = \lambda \vec{a} \implies \vec{r} = \vec{b} + \lambda \vec{a}$$

$$(\vec{b} + \lambda \vec{a}) \cdot \vec{a} = 0$$

$$\vec{b} \cdot \vec{a} + \lambda (\vec{a} \cdot \vec{a}) = 0 \implies \vec{b} \cdot \vec{a} + \lambda |\vec{a}|^2 = 0$$

$$\lambda = -\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}$$

$$\vec{b} \cdot \vec{a} = (0)(\sqrt{7}) + (1)(1) + (2)(-1) = 0 + 1 - 2 = -1$$

$$| \vec{a} |^2 = (\sqrt{7})^2 + (1)^2 + (-1)^2 = 7 + 1 + 1 = 9$$

$$\lambda = -\frac{-1}{9} = \frac{1}{9}$$

$$|3\vec{r}|^2 = 9 |\vec{r}|^2$$

$$|\vec{r}|^2 = \vec{r} \cdot \vec{r} = \left(\vec{b} + \frac{1}{9}\vec{a}\right) \cdot \left(\vec{b} + \frac{1}{9}\vec{a}\right)$$

$$|\vec{r}|^2 = |\vec{b}|^2 + \frac{2}{9}(\vec{b} \cdot \vec{a}) + \frac{1}{81}|\vec{a}|^2$$

$$|\vec{b}|^2 = 0^2 + 1^2 + 2^2 = 1 + 4 = 5$$

$$|\vec{r}|^2 = 5 + \frac{2}{9}(-1) + \frac{1}{81}(9)$$

$$|3\vec{r}|^2 = 9 \left(\frac{44}{9}\right) = 44$$