Calculate Magnitude Squared of Vector Satisfying Cross and Dot Product Conditions

Given the vectors: $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ $\vec{b} = \hat{j} + 2\hat{k}$

We are given the relation: $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$

Using the property of cross products $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$, we can rewrite the equation as: $\vec{r} \times \vec{a} - \vec{b} \times \vec{a} = \vec{0}$ $(\vec{r} - \vec{b}) \times \vec{a} = \vec{0}$

This implies that the vector $(\vec{r} - \vec{b})$ is collinear with $\vec{a}$. Therefore, there exists a real scalar $\lambda$ such that: $\vec{r} - \vec{b} = \lambda \vec{a} \implies \vec{r} = \vec{b} + \lambda \vec{a}$

Next, we use the second condition, $\vec{r} \cdot \vec{a} = 0$: $(\vec{b} + \lambda \vec{a}) \cdot \vec{a} = 0$ $\vec{b} \cdot \vec{a} + \lambda |\vec{a}|^2 = 0$ $\lambda = -\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}$

Now, let's calculate the magnitudes and dot products: $|\vec{a}|^2 = (\sqrt{7})^2 + (1)^2 + (-1)^2 = 7 + 1 + 1 = 9$ $\vec{a} \cdot \vec{b} = (\sqrt{7})(0) + (1)(1) + (-1)(2) = 0 + 1 - 2 = -1$

Substituting these values into the expression for $\lambda$: $\lambda = -\frac{-1}{9} = \frac{1}{9}$

So, the vector $\vec{r}$ is: $\vec{r} = \vec{b} + \frac{1}{9}\vec{a}$

Multiplying both sides by $3$: $3\vec{r} = 3\vec{b} + \frac{1}{3}\vec{a}$

Substituting the explicit components of $\vec{a}$ and $\vec{b}$: $3\vec{r} = 3(\hat{j} + 2\hat{k}) + \frac{1}{3}(\sqrt{7}\hat{i} + \hat{j} - \hat{k})$ $3\vec{r} = \frac{\sqrt{7}}{3}\hat{i} + \left(3 + \frac{1}{3}\right)\hat{j} + \left(6 - \frac{1}{3}\right)\hat{k}$ $3\vec{r} = \frac{\sqrt{7}}{3}\hat{i} + \frac{10}{3}\hat{j} + \frac{17}{3}\hat{k}$

Now, calculating the magnitude squared $|3\vec{r}|^2$: $|3\vec{r}|^2 = \left(\frac{\sqrt{7}}{3}\right)^2 + \left(\frac{10}{3}\right)^2 + \left(\frac{17}{3}\right)^2$ $|3\vec{r}|^2 = \frac{7}{9} + \frac{100}{9} + \frac{289}{9}$ $|3\vec{r}|^2 = \frac{7 + 100 + 289}{9} = \frac{396}{9} = 44$

Hence, $|3\vec{r}|^2$ is equal to $44$.