Intersection between 2 planes I need help understanding the formula for intersection between 2 planes 

Where is $\alpha$ actually, its not clear in the diagram. It appears that I could just interpret it as the below?

Where $\beta = \theta$?
UPDATE
Now given the question: 

What I did

Whats wrong? Correct answer $60.7^{\circ}$. Did I "expanded" the equation for plane whose equation was in Cartesian form correctly?
 A: The norm $\vec{n}=\langle 0,3,5 \rangle$ is $\sqrt{34}$ and not $\sqrt{24}$. Then, the computation adds up and the angle amounts to 60.7$^\circ$.
A: 
Maybe it's best to ignore the diagram from your book. Just consider two planes with normal vectors ${\bf n}_1$ and ${\bf n}_2$. Let's suppose that the angle between these vectors is $\alpha$ (which is acute). Then that's the angle between the two planes. 
Now what if our first plane had had the normal vector $-{\bf n}_1$? Then the angle between our two normals would have been $\theta$. This angle is obtuse. So it should be replaced by $\alpha = \pi - \theta$. 
Anytime you are computing the angle between two planes (which is done by computing the angle between their normal vectors), you should make sure your answer is an acute angle. If your initial answer is obtuse, you've used normals which point out of "incompatible" sides of the planes. Swapping one of the normals for its negative will change the angle to its complement and then give you an acute angle (the desired answer).
