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 $\vec{l_1}(a,b,c)$.
Thus, $a-b+c=0$ and $2a+b-c=0$, which gives $a=0$ and $b=c$ and $\vec{l_1}(0,1,1).$
Let $\vec{l_2}(a,b,c)$.
Thus, $a+b+c=0$ and $2a+3b-c=0$, which gives $b=-\frac{3}{4}a$, $c=-\frac{1}{4}a$ 
and $\vec{l_1}(4,-3,-1).$
Now,
$$\measuredangle\left(L_1,L_2\right)=\arccos\frac{|0\cdot4+1\cdot(-3)+1\cdot(-1)|}{\sqrt{4^2+(-3)^2+(-1)^2}\sqrt{0^2+1^2+1^2}}=\arccos\frac{2}{\sqrt{13}}.$$
