Find the equation of the line $ r $ Find the equation of the line $ r $ that goes through $ (1, -2,3) $, concurrent with the line
$$\begin{cases}
x=2+3t \\
y=1+2t \\
z= -1 \\
\end{cases}$$
and has orthogonal director vector $ (1, -3,1) $
Solution: line r is contained in the plane $(x-1)-3(y+2)+(z-3)=0$
$x-3y+z=10$
Next we find the intersection of the plane and the concurrent line
$2+3t-3(1+2t)-1=10$
$-3t-2=10$
$t=-4, so (-10,-7,-1)$
So line r goes through the points (1,-2,3) and (-10,-7,-1)
$\begin{cases}
x=1-11t \\
y=-2-5t \\
z=3-4t \\
\end{cases}$
How to solve without using plan concepts? (Whatever if it's parametric, symmetrical, vector representation ....)
 A: Here is another approach:
Let $\vec r=(a,b,c)$ be a directing  vector of the line we search. Its parametric equation is then given by
$$\begin{cases}
x=\phantom{-}1+au,\\ y=-2+bu, \\ z=\phantom{-}3+cu,\end{cases}\quad(u\in\mathbf R).$$
It meets the given line if and only if the linear system (in $t$ and $u$):
$$\begin{cases}
\phantom{-}1+au=2+3t, \\
-2+bu=1+2t, \\
\phantom{-}3+cu= -1,\\
\end{cases}\iff \begin{cases}
au-3t=1, \\
 bu-2t=3, \\
cu= -4,\\
\end{cases}$$
has a solution.
Note the last equation implies that $u, c\ne 0$. Also, a directing vector is defined up to a nonzero scalar multiple, so we may choose the value of $c$ to simplify the computation. We'll take $c=4$, so $u=-1$. 
Now the orthogonality condition  is
$$a-3b-4=0,$$
whence a linear system for $a$ and $b$
\begin{cases}
a+3t=-1, \\
b+2t=-3, \\
a-3b=4,\\
\end{cases}
which can easily be solved with RREF of the augmented matrix.
A: If concurrent means intersecting (which to me is confusing) then your solution is fine, I think. The orthogonal plane is correct and the intersecting point too. The parametrised line indeed lies in that plane etc. 
I see no issues with it otherwise.
A: Use the information that you’ve been given directly. The two lines are coincident, so a direction vector of the line that you’re trying to find is $(2+3t,1+2t,-1)-(1,-2,3)$ for some value of $t$. This vector must be orthogonal to $(1,-3,1)$, so set their dot product equal to zero and solve for $t$.
A: A straight $r$ going through point $(1,-2,3)$ has this parametric form:
$$x = 1 + a·u$$
$$y = -2 + b·u$$
$$z= 3 + c·u$$
If it has intersection with the other given line, then
$$ x = 1 + au = 2+ 3t$$
$$ y= -2 + bu = 1 +2t$$
$$ z= 3 + cu = -1$$
If it has a orthogonal director vector (1,−3,1) then the dot product with $(a,b,c)$ must be $0$
$$ 1·a -3·b + 1·c = 0$$
A: This approach ends up computing the plane equation implicitly anyways:
Let's write down all we have:
Point: $\mathbf{p}=(1,-2,3)^T$
Line: $\mathbf{x}=\mathbf{a}+t\mathbf{l}_1,\,\mathbf{a}=(2,2,-1)^T,\,
\mathbf{l}_1=(3,2,0)^T$
The normal vector:  $\mathbf{n}=(1,-3,1)^T$
Thus, $\mathbf{x}=\mathbf{p}+v\mathbf{l}_2,\,
\mathbf{l}_2\mathbf{n}=0$ and $\mathbf{p}+v\mathbf{l}_2=\mathbf{a}+t\mathbf{l}_1$ is the point.
Now, let's use $\mathbf{l}_2\mathbf{n}=0$ by multiplying
$\mathbf{p}+v\mathbf{l}_2=\mathbf{a}+t\mathbf{l}_1$ by $\mathbf{n}$ to get rid of $v\mathbf{l}_2$ -- that's the same thing with the plane.
$(\mathbf{p}-\mathbf{a}).\mathbf{n}=t\mathbf{l}_1.\mathbf{n}$
$t=\frac{(\mathbf{p}-\mathbf{a}).\mathbf{n}}{\mathbf{l}_1.\mathbf{n}}$ $=-5$.
So we've got the 2nd point at desired line: $\mathbf{p}_2=\mathbf{a}+\frac{(\mathbf{p}-\mathbf{a}).\mathbf{n}}{\mathbf{l}_1.\mathbf{n}}\mathbf{l}_1$ $=(-13, -8, -1)^T$ and can write the line as $\mathbf{x}=\mathbf{p}+u(\mathbf{p}_2-\mathbf{p})$.
Can you take it from here?
Note, however, $t=-5$ and not $-4$ as in your solution, so feel free to correct me if I'm wrong somewhere.
