I'm at my wit's end with this problem: Find a parametric representation of the line $\ell$ which passes through the point $(3,2,-1)$ and intersects the lines $$\ell_1: \left\{ \begin{alignedat}{3} x & ={} & 1 +{} &s \\ y & ={} & &s \\ z & ={} & -1+{} &s \end{alignedat} \right. \qquad \ell_2: \left\{ \begin{alignedat}{3} x & ={} &10 +{} & 5&t \\ y & ={} &5 +{} & &t \\ z & ={} &2 +{} & 2&t \end{alignedat} \right.\tag*{($s,t \in \mathbf{R}$)} $$
I feel like this seems like a typical routine problem that shouldn't stump me like this, but I can't for the life of me figure out a nice solution.
I've tried the following:
Given an orthonormal system $O\mathbf{e}_1\mathbf{e}_2\mathbf{e}_3$ and $$ \begin{align*} P:(3,2,-1) && Q:(1+s,s,-1+s) && R: (10+5t,5+t,2+2t), \end{align*} $$ we want to find a direction vector $\overrightarrow{QR}$ such that $$ \overrightarrow{QR} = \overrightarrow{QP} + \overrightarrow{PR} = (\overrightarrow{OP} - \overrightarrow{OQ}) + (\overrightarrow{OR} - \overrightarrow{OP}) = \overrightarrow{OR}-\overrightarrow{OQ}. $$ We want this vector to be parallel to $\overrightarrow{QP} = \overrightarrow{OP} - \overrightarrow{OQ}$ so that it hits all 3 points. Hence $$ \overrightarrow{QR} = k \overrightarrow{QP}\tag*{($k\in\mathbf{R}$)} $$ which gives us \begin{align*} \overrightarrow{QR}=\overrightarrow{OR}-\overrightarrow{OQ} &= (10+5t-1-s,5+t-s,2+2t+1-s) = \\ &=(9+5t-s,5+t-s,3+2t-s) = k(2-s,2-s,-s) \end{align*} iff $$ \left\{ \begin{alignedat}{4} 9 & +{} & 5t-s & &{} ={} & &{} 2k-ks \\ 5 & +{} & t-s & &{} ={} & &{} 2k-ks \\ 3 & +{} & 2t-s & &{} ={} & &{} -ks \end{alignedat} \right. $$ and I have no clue how to solve a system that looks like this, but solving it in Mathematica gives me the parameters $s=-2$ and $t=-1$.We then have \begin{align*} Q(-2): (-1,-2,-3) && R(-1):(5,4,0), \end{align*} and given an arbitrary point $S: (x,y,z)$ on the line and the coordinates for the direction vector, we can then find the line's parametric equation with the following equation: \begin{align*} \overrightarrow{QS}&=\lambda\overrightarrow{QR}=\lambda(\overrightarrow{OR} - \overrightarrow{OQ}) =\lambda(6,6,3) =\\ &=\overrightarrow{OS} - \overrightarrow{OQ}=(x+1,y+2,z+3) \end{align*} hence the line $$\ell_3: \left\{ \begin{alignedat}{3} x & ={} &-1 +{} & 6&\lambda \\ y & ={} &-2 +{} & 6&\lambda \\ z & ={} &-3 +{} & 3&\lambda \end{alignedat} \right.\tag*{($\lambda \in \mathbf{R}$)} $$ This solution looks very "all over the place" and I'm not even confident in the steps or the solution, but I've verified it in Mathematica and it is a solution.
The conclusion is that I'm looking for a neater solution and some insight to the problem. I've played around with the lines $\ell_1$ and $\ell_2$ and the plane that they span with the point $P$, but I don't really know how to proceed from there. My intuition tells me that I have to use that plane in some way.
First, consider the plane $\Pi_1$ formed by $\ell$ and $\ell_1$. The vector $[1,1,1]$ is is along the line $\ell_1$ and thus in the plane $\Pi_1$. Putting $s=0$, the point $(1,0,-1)$ is on $\ell_1$ and hence on $\Pi_1$. The point $(3,2,-1)$ i s also in $\Pi_1.$ Subtracting coordinates, we see that the vector $[2,2,0]$ lies in the plane $\Pi_1$ so the cross-product $n_1=[1,1,1] \times [2,2,0]$ is normal to $\Pi_1$ and thus perpendicular to a vector in the direction of $\ell.$ Similarly, we can consider the plane $\Pi_2$ formed by $\ell$ and $\ell_2$ and find a vector $n_2$ normal to $\Pi_2$ and hence perpendicular to a vector in the direction of $\ell.$ Then the cross-product $n_1 \times n_2$ is in the direction of $\ell.$ Since $(3,2,-1)$ is on $\ell$, we can immediately write the parametric equation for $\ell$.