Find the intersection points of a line parallel to a given line with two lines Given:
$$Line\ l \leftrightarrow \frac{2x-3}{2} = \frac {y}{2}=\frac{2z+1}{6}$$
$$Line\ m \leftrightarrow 3x-5=0 \vert 3y+3z+2=0$$
Question find the point L on l and the point M on m such that LM is parallel to p with following equation:
$$p \leftrightarrow x= \frac {y}{6}=\frac {z}{2}$$
I solved this by chosing two points $L (x1,y1,z1)$ and $M (x2,y2,z2)$. I know that $M-L // p$. So i have the following system of equations:
$$\frac{5}{3}-(\lambda+\frac{3}{2})=k$$
$$\mu-2\lambda=6k$$
$$-\mu-\frac{2}{3}-(3\lambda-\frac{1}{2})=2k$$
And then I filled in the parametric equation of m for M and that of l for L. 
As a solution I got $L (9/5;3/5;-1/5)$ and $M (5/3;-1/5;-7/15)$. Is this correct or did I do something wrong?
 A: One way to start is by finding parametric equations for $l$ and $m$. For the first line, this can be done by setting the given equation equal to $\lambda$ and then solving for $x$, $y$ and $z$. Doing this should give you something like $$L=\lambda\left(1,2,3\right)+\left(\frac32,0,-\frac12\right).$$  
For line $m$, one way to find a parameterization is to switch to homogeneous coordinates. The line is essentially given as the intersection of two planes. If we homogenize those equations, then the line can be described as the null space of the matrix $$\begin{bmatrix}3&0&0&-5\\0&3&3&2\end{bmatrix}.$$ This matrix is already in echelon form, so one can read from it that its null space is spanned by $[5,-2,0,3]^T$ and $[0,-1,1,0]^T$. Converting that back into inhomogeneous Cartesian coordinates yields $$M=\mu\left(0,-1,1\right)+\left(\frac53,-\frac23,0\right).$$ If there’s another way of finding this parametric equation that makes more sense to you, use it, but you should end up with something equivalent.  
Now, make use of the condition that $(L-M)$ is parallel to $p$. Taking advantage of the fact that you’re working in $\mathbb R^3$, this condition can be expressed as $(L-M)\times \mathbf p=0$, where $\mathbf p$ is the line’s direction vector. If you expand this out using the above parameterizations of $L$ and $M$, you’ll end up with a system of linear equations in $\lambda$ and $\mu$, which I expect you know how to solve.
