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Let the given vector be $\vec{a} = \hat{i} + \hat{j} + \hat{k}$.
Find the magnitude of $\vec{a}$: $|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
Find the unit vector along $\vec{a}$: $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$.
We need to find vectors of magnitude $3\sqrt{3}$ which are collinear to $\vec{a}$. These vectors will be of the form $\pm 3\sqrt{3} \hat{a}$.
So, the required vectors are $\pm 3\sqrt{3} \left( \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \right) = \pm (3\hat{i} + 3\hat{j} + 3\hat{k})$.
Therefore, the vectors are $3\hat{i} + 3\hat{j} + 3\hat{k}$ and $-3\hat{i} - 3\hat{j} - 3\hat{k}$.
Correct Answer: $3\hat{i} + 3\hat{j} + 3\hat{k}$ and $-3\hat{i} - 3\hat{j} - 3\hat{k}$