Parallel Vectors

$$\vec{r} \times \vec{a} - \vec{b} \times \vec{a} = \vec{0} \implies (\vec{r} - \vec{b}) \times \vec{a} = \vec{0}$$
This implies that the vector $$(\vec{r} - \vec{b})$$ is parallel to $$\vec{a}$$. Therefore, we can write:
$$\vec{r} - \vec{b} = \lambda \vec{a} \implies \vec{r} = \vec{b} + \lambda \vec{a}$$
Now, substitute this into the second condition ($$\vec{r} \cdot \vec{b} = 0$$):
$$(\vec{b} + \lambda \vec{a}) \cdot \vec{b} = 0$$
$$|\vec{b}|^2 + \lambda (\vec{a} \cdot \vec{b}) = 0$$
Let's compute the scalar values:
$$|\vec{b}|^2 = 1^2 + (-1)^2 + 1^2 = 3$$
$$\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (1)(1) = 1$$
Substitute these back to solve for $$\lambda$$:
$$3 + \lambda(1) = 0 \implies \lambda = -3$$
Now, find vector $$\vec{r}$$:
$$\vec{r} = \vec{b} - 3\vec{a} = (\hat{i} - \hat{j} + \hat{k}) - 3(\hat{i} + \hat{j} + \hat{k}) = -2\hat{i} - 4\hat{j} - 2\hat{k}$$
Finally, compute $$|\vec{r}|^2$$:
$$|\vec{r}|^2 = (-2)^2 + (-4)^2 + (-2)^2 = 4 + 16 + 4 = 24$$