let there be a line L: $\frac{x-1}{2}= \frac{y+1}{3}= \frac{z}{1}$ and a plane: $2x-y-z=5$. With this given data find: a line L1, such that L1 is parallel to L, is in P, and the distance between L and L1 is d.
I did manage to find d , which is $\frac{2}{\sqrt{6}}$. I have no ideas as to how to continue this however.
First, let us write the line in an easier (for me, at least) way:
$$3x-3=2y+2=6z\implies\begin{cases}x=2z+1\\{}\\y=3z-1\end{cases}\;\implies\;\ell:\;(1,-1,0)+t(2,3,1)\;,\;\;t\in\Bbb R$$
As it should be because of the problem requirements, check we have that $\;\ell\parallel\pi\;$. A parametric representation for the plane could be:
$$\begin{cases}A=(3,1,0)\\B=(2,0,-1)\\C=(0,-4-1)\end{cases}\in\pi\implies\pi:\;B+t\vec{BA}+s\vec{BC}\;,\;\;t,s\in\Bbb R\implies$$
$$\pi:\;(2,0,-1)+t(1,1,1)+s(-2,-4,0)$$
The wanted distance between $\;\ell,\,\pi\;$ is the distance of any point on the line to the plane, so for example let us take the point $\;(1,-1,0)\;$ so its distance to $\;\pi\;$ is:
$$d=\frac{|2+1+0-5|}{\sqrt6}=\frac2{\sqrt6}=\sqrt\frac23$$
Try to take it from here. The continuation may now depend heavily, in my opinion, in what you know/prefer. You could take any point on the line $\;\ell\;$ and look at its orthogonal projection on the plane $\;\pi\;$ and then choose the line on the plane thorugh the foot of that projection on the plane, or you could set an equation for a general point in the plane which is at distance $\;d\;$ from say $\;(1,-1,0)\in\ell\;$ and solve with differential calculus.