Angle between planes with x, y and z variables Problem: Find the angle between the planes $3x-y+z-5=0$ and $x+2y+2z+2=0$.
We have been thought 1 formula for solving angles which is : Angle $= Arctan(\frac{m1-m2}{1+m1m2})$ but that is for parallel lines only and for equation with x and y only. 
How to answer this kind of problem? Thank you.

Finding the angle between two planes is similar to finding the angle between two vector (normal of the planes). The first plane is $3x-y+z-5=0$ has normal  $\overrightarrow n_0 =\langle 3, -1, 1 \rangle$. The second plane is $x+2y+2z+2=0$ has normal  $\overrightarrow n_1 =\langle 1, 2, 2 \rangle$.
To find the angle between two vectors we use the formula
$$cos(\theta)= \frac{\mathbf A \bullet \mathbf B}{|\mathbf A||\mathbf B|}$$
Insert A as $\overrightarrow n_0$ and B as $\overrightarrow n_1$.
$$\overrightarrow n_0 \bullet \overrightarrow n_1=3*1+(-1)*(2)+1(2) = 3$$
$$|\overrightarrow n_0| = \sqrt{3^2+(-1)^2+1^2}=\sqrt{11}$$
$$|\overrightarrow n_1| = \sqrt{1^2+2^2+2^2}=3$$
So,
$$cos(\theta)= \frac{3}{\sqrt{11}*3}=\frac{1}{\sqrt{11}}$$
Therefore
$$\theta= \arccos(\frac{1}{\sqrt{11}}) \approx 1.26451896 rad$$
Relevant Find the angle between two planes using their normal vectors
