In homogeneous coordinates we have the point $M(1, 0, 2, 1)$ and the line $$a:\left\{ \begin{array}{c} x=2\lambda+3\mu \\ y=1\lambda-2\mu\\ z=-1\lambda-\mu\\ t=3\lambda+\mu \end{array} \right. $$
I have to find equation of a line $b$ which is parallel to the line $a$ and passes the point $M$.
My understanding is the following. Since the lines $a$ and $b$ are parallel, they have common infinite point, so I find the coordinates of an infinite point using the coefficients in front of the $\mu$, i.e. the point P(3, -2, -1, 0).
Having two points, I form the equation of the line $b$, i.e. $$b:\left\{ \begin{array}{c} x=1\lambda+3\mu \\ y=0\lambda-2\mu\\ z=2\lambda-\mu\\ t=1\lambda+0\mu \end{array} \right. $$
Is this correct?
Thanks in advance!
What you did with “using the coefficients in front of the $\mu$” was picking $\lambda=0,\mu=1$ which leads to $P(3,-2,-1,1)$ which is not at infinity!
The point at infinity is characterized by $t=3\lambda+\mu=0$ so for example $\lambda=1,\mu=-3$ (but any other multiple of this would work, too). With that you get $P(-7,7,2,0)$. Taking this correction into account, the rest of your approach looks good.