The acute angle between the planes $2x- y + z = 6 $ and $ x + y = 2 z = 3, $ is

60°

Vectors normals to the given planes are $\vec{n_1}= 2\hat{i} - \hat{j} + \hat{k} $ and $ \vec{n_2} = \hat{i}+\hat{j} + 2\hat{k}.$

Let $\theta $ be the acute angle between the planes. Then, 

$cos \theta = \frac{|\vec{n_1}.\vec{n_2}|}{|\vec{n_1}||\vec{n_2}|}=\frac{|2-1+z|}{\sqrt{6}\sqrt{6}}=\frac{1}{2}⇒ \theta = 60°$