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 ....)

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 $t$.
