Decide the line through two points I have this problem below that i don't understand:

Let $L_1$ be a straight line in $\mathbb{R}^3$ as defined by $(x, y, z) = (2,2,0) + t(3,0,2)$.


a) Determine the plane that contains the line $L_1$ and the point $A = (8, 2, 3)$.

And i solved a with taking the point in the line $L_1$ with $(8,2,3)-(2,2,0) = (6,0,3)$.
And after that i did the Cross-product on the vector $(3,0,2) \times (6,0,3)$ and i got the
vector that is ortogonal on these two vectors $(0,3,0)$.
And after that i put in the point $(2,2,0)$ in the plane equation and got that the plane is
$y=2$.
But then i came to the b) part that troubled me below:

b) The line $L_2$ is defined by $(x, y, z) = (5, 1, 0) + t (2, 1, 1)$. Determine an equation for that line which passes through the point $A = (8, 2, 3)$ and intersects both $L_1$ and $L_2$.

I don't know what to begin to right here they say you should use the plane equation you
got in part a) with is $y=2$ but i'don't understand.
 A: The line you are looking for passes through $A$ and it intersects $L_1$. Therefore it is along the plane $y=2$, which contains $A$ and  any point in the line $L_1$. It follows that such line should pass through the point $B$ given by the intersection of $y=2$ and $L_2$:
$$(x, 2, z) = (5, 1, 0) + t (2, 1, 1) \implies t=1, \quad
B=(5, 1, 0) + 1 (2, 1, 1)=(7,2,1).$$
So this line goes through $A=(8,2,3)$ and $B=(7,2,1)$. Can you finish the job?
A: The answer by Robert Z is fine and I cannot improve it, but I will provide my answer nevertheless, since it is about how to solve part b) without using part a) of the problem.
An arbitrary point of $L_1$ has the form $(2,2,0)+t(3,0,2)$. The line defined by $A$ and that point is the line$$\left\{(8,2,3)+u\bigl((2,2,0)+t(3,0,2)-(8,2,3)\bigr)\,\middle|\,u\in\Bbb R\right\}$$which is equal to$$\left\{(8,2,3)+u\bigl((-6,0,-3)+t(3,0,2)\bigr)\,\middle|\,u\in\Bbb R\right\}.$$A point of this line belongs to $L_2$ if it is of the form $(5,1,0)+v(2,1,1)$, for some $v\in\Bbb R$. So, you want to know when is it that we have$$(8,2,3)+u\bigl((-6,0,-3)+t(3,0,2)\bigr)=(5,1,0)+v(2,1,1).$$In other words, when do we have$$\left\{\begin{array}{l}3 t u-6 u-2 v+3=0\\1-v=0\\2 t u-3 u-v+3=0?\end{array}\right.$$It follows from the second equation that $v=1$ and then $(5,1,0)+v(2,1,1)=(7,2,1)$. So, the line that you're after is the line defined by $A$ and $(7,2,1)$.
