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 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 $0$
$$ 1·a -3·b + 1·c = 0$$
