Find the parametric representation of a line 
Find the parametric representation of a line $l$ so that $l$ is in $\pi:\overline X=(3,0,0)+t(2,3,1)+s(6,-2,3)$ and $l$ intersects $l_1:\overline X=(6,3,0)+t(1,1,2)$ with a 60° angle.

I did find the intersection point at $(7,4,2)$ by equating the coordinates of said point expressed once as a point on $\pi$ and again as a point on $l_1$.
I tried using the dot product, since it has $\cos\theta$ as the angle between the two vectors, but I get stuck since I don't know the length of any of the vectors. I'm not even sure I need their length. I'd love some help please.
 A: First of all, rewrite $\pi$ in normal form:
$$(2,3,1)×(6,-2,3)=(11,0,-22)=11(1,0,-2)$$
$$(3,0,0)\cdot(1,0,-2)=3$$
$$\pi:\mathbf r\cdot(1,0,-2)=3$$
The equation of $l$ is $(7,4,2)+\lambda\mathbf b$ where $\lambda\in\Bbb R$ and $\mathbf b$ is a vector with the following properties:


*

*It is perpendicular to $\pi$'s normal: $\mathbf b\cdot(1,0,-2)=0$. Writing $\mathbf b=(a,b,c)$ we have
$$a=2c\tag1$$

*It makes a 60° angle with $l_1$:
$$\mathbf b\cdot(1,1,2)=\sqrt{1^2+1^2+2^2}\sqrt{a^2+b^2+c^2}\cos60^\circ$$
$$a+b+2c=\frac{\sqrt6}2\sqrt{a^2+b^2+c^2}\tag2$$


Now assume $\sqrt{a^2+b^2+c^2}=a^2+b^2+c^2=1$, i.e. $\mathbf b$ is a unit vector. (2) then simplifies, and (1) removes $a$ from it:
$$b+4c=\frac{\sqrt6}2$$
$$b=\frac{\sqrt6}2-4c\tag3$$
$b^2$ can be written in terms of $c$:
$$a^2+b^2+c^2=1$$
$$(2c)^2+b^2+c^2=5c^2+b^2=1$$
$$b^2=1-5c^2$$
Squaring (3) we get
$$1-5c^2=16c^2-4\sqrt6c+\frac32$$
$$21c^2-4\sqrt6c+\frac12=0$$
Solving for $c$:
$$c=\frac{4\sqrt6\pm\sqrt{54}}{42}=\frac{\sqrt6}6\text{ or }\frac{\sqrt6}{42}$$
From here two solutions for $\mathbf b$ may be constructed:
$$\mathbf b=\left(\frac{\sqrt6}3,-\frac{\sqrt6}6,\frac{\sqrt6}6\right)\text{ or }\left(\frac{\sqrt6}{21},\frac{17\sqrt6}{42},\frac{\sqrt6}{42}\right)$$
Removing the factor of $\sqrt6$ and common denominators leads us to the two solutions for $l$:
$$l=(7,4,2)+\lambda(2,-1,1)\text{ or }(7,4,2)+\lambda(2,17,1),\ \lambda\in\Bbb R$$
