To find the angle

\(\theta\) between two vectors \(\vec{A}\) and \(\vec{B}\): Observe that: \[ \vec{B} = 6\hat{i} - 3\hat{j} + 6\hat{k} = 3(2\hat{i} - \hat{j} + 2\hat{k}) = 3\vec{A} \]

Since

\(\vec{B}\) is a positive scalar multiple of \(\vec{A}\) (\(\vec{B} = k\vec{A}\) where \(k = 3 \gt 0\)), the two vectors point in the exact same direction and are collinear/parallel. Alternatively, using the dot product formula: \[ \cos\theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} = \frac{(2)(6) + (-1)(-3) + (2)(6)}{\sqrt{2^2 + (-1)^2 + 2^2} \cdot \sqrt{6^2 + (-3)^2 + 6^2}} = \frac{12 + 3 + 12}{\sqrt{9} \cdot \sqrt{81}} = \frac{27}{3 \times 9} = \frac{27}{27} = 1 \] \[ \theta = \cos^{-1}(1) = 0^\circ \]