$$ \text{Problem: Find } \mathbf{r}(t) \text{ of a particle starting at } (1, -1, 2) \text{ moving toward } (3, 0, 3), \ \text{with } \mathbf{a} = 2\mathbf{i} + \mathbf{j} + \mathbf{k} \text{ and initial speed 2 at } t = 0. \\ \text{Flawed Reasoning: Assumed } \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2, \ \mathbf{r}_0 = (1, -1, 2), \ \mathbf{v}_0 = 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}, \\ \text{so } \mathbf{r}(t) = (1 + 2t + t^2)\mathbf{i} + (-1 + 2t + \frac{1}{2}t^2)\mathbf{j} + (2 + 2t + \frac{1}{2}t^2)\mathbf{k}. \\ \text{Why Wrong: Speed } |\mathbf{v}_0| = \sqrt{12} \approx 3.46 \neq 2, \ \text{and } \mathbf{v}_0 = 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \text{ not parallel to direction } (2, 1, 1). \\ \text{Correct Approach: } \mathbf{v}(t) = (2t + c_1)\mathbf{i} + (t + c_2)\mathbf{j} + (t + c_3)\mathbf{k}, \ |\mathbf{v}(0)| = 2, \ \text{direction } (2, 1, 1), \\ (c_1, c_2, c_3) = k(2, 1, 1), \ \sqrt{6k^2} = 2, \ k = \frac{\sqrt{6}}{3}, \ \mathbf{v}(0) = \frac{2\sqrt{6}}{3}\mathbf{i} + \frac{\sqrt{6}}{3}\mathbf{j} + \frac{\sqrt{6}}{3}\mathbf{k}, \\ \mathbf{r}(t) = \left(t^2 + \frac{2\sqrt{6}}{3}t + 1\right)\mathbf{i} + \left(\frac{1}{2}t^2 + \frac{\sqrt{6}}{3}t - 1\right)\mathbf{j} + \left(\frac{1}{2}t^2 + \frac{\sqrt{6}}{3}t + 2\right)\mathbf{k}, \\ \text{Takeaway: Speed is magnitude; velocity aligns with } (2, 1, 1); \ \text{integrate with conditions.} $$