How to find the angle between 2 intersection lines of four planes? 
Given $L_1=\begin{cases}P_1: x-y+z=4\\ P_2: 2x+y-z=6 \end{cases}$ 
and $L_2=\begin{cases}P_1:x+y+z=4\\ P_2:2x+3y-z=6 \end{cases}$
$L_1$ and $L_2$ are the intersection lines of the given planes. Find the angle between the lines $L_1,L_2$

First in order to find the intersection lines between the respective planes it is enough to find to dots that belong to the intersection line, so for $L_1$:
$$
P_1+P_2 \Rightarrow 3x=10 \Rightarrow x={10 \over 3}
$$
Then: 
$$
2 \cdot \frac{10}{3}+y-z=6 \Rightarrow y-z=-\frac{2}{3}
$$
For $z=0$ and $z=1$ we get $y=-\frac{2}{3}$ and $y=\frac{1}{3}$ respectively so now can represent $L_1$ parametrically: $(\frac{10}{3},\frac{1}{3}, 10)+t(0,-1,0)$ where $t(0,-1,0)$ is the direction vector. 
After identical calculations we can find $L_2: (0,\frac{10}{4}, \frac{3}{2})+t(\frac{10}{3}, 0,-\frac{2}{3})$.
Now we can calculate the angle of $L_1, L_2$ via scalar product:
$$
\cos \theta=\frac{L_1 \cdot L_2}{\Vert{L_1}\Vert \cdot \Vert L_2 \Vert}=0 \Rightarrow \theta=90^{\circ}
$$
I'm not sure that my answer is correct and I feel I'm missing something.

Let me answer the general case. 
First finding intersection of planes:
In general planes in $\mathbb{R}^3$ are of the form:
$$ax+by+cz+d=0.$$
The intersection of two planes $ax+by+cz+d=0$ and $Ax+By+Cz+D=0$ exists if and only if $|aA+bB+cC|\neq ||(a,b,c)||.||(A,B,C)||$ or if $|aA+bB+cC|=||(a,b,c)||.||(A,B,C)||$, then $dC=Dc$. The latter case means planes coincide.
Note that $|aA+bB+cC|=||(a,b,c)||.||(A,B,C)||$ implies two planes are parallel.
If two planes have intersection then by letting one of the variables $x$, $y$ or $z$ equal to $t$ we will find other variables. Here let $z=t$ then solve:
$$\left\{\begin{array}{} ax+by=d-ct\\ Ax+By=D-Ct\end{array}\right.$$
we are now able to find values of $x$ and $y$. After calculation we have something like the following form:
$$\frac{x-a'}{a''}=\frac{y-b'}{b''}=\frac{z}{1}(=t).$$
Now, do the similar arguments for two other planes. We have:
$$\frac{x-A'}{A''}=\frac{y-B'}{B''}=\frac{z}{1}(=T).$$
Now we have two vectors $(A'',B'',1)$ and $(a'',b'',1)$ and by using the formula:
$a''A''+b''B''+1=\sqrt{a''^2+b''^2+1}.\sqrt{A''^2+B''^2+1}.cos(\alpha)$ the angle $\alpha$ is easy to obtain.
Now in your question: $$L_1: x=\frac{10}{3},y+\frac{2}{3}=z(=t)$$ here $v_1=(0,1,1)$ and $$L_2:\frac{x-6}{-4}=\frac{y+2}{3}=z(=t)$$ here $w=(-4,3,1)$. So, $$cos(\alpha)=\frac{4}{\sqrt{2}\sqrt{26}}=\frac{2\sqrt{13}}{13}$$
