Line that intersects three other lines in $\mathbb A^4$ I'm studying for my affine geometry test and I'm struggling with the following exercise.
Given $r_1:x_1=x_2=x_3=x_4, r_2=(0,0,0,1)+\langle (1,0,0,0) \rangle, r_3=(0,1,0,0)+\langle(0,0,a,1)\rangle$ with $a \in \mathbb R$. Determine if there exists an $a \in \mathbb R$  for which we can find a line such that intersects $r_1,r_2,r_3$. Is it unique?
I have already worked out that $r_1 \cap r_2 = \emptyset$ and $D(r_1) \cap D(r_2)= \{0\}$, where $D(r_i)$ denotes the vector space of the respective affines subspace. It is also easy to see that the same conditions follow with $r_3$; $r_1,r_2,r_3$ are lines which don't intersect and the insertection of the vector spaces of two of the lines is the trivial vector space. My reasoning now is the following:
Suppose there exists $r \subset \mathbb A^4(\mathbb R)$ such that intersects $r_1,r_2,r_3$. Then, it follows that there exist $P_1 \in r_1,P_2 \in r_2$ such that $P_1 \in r,P_2 \in r$. Thus, $D(r)= \langle \vec{P_1 P_2} \rangle$ and $P_1 \in r$, hence it follows that $r \subset r_1+r_2$, which I calculated $r_1 + r_2 = (0,0,0,1)+ \langle (1,0,0,0),(1,1,1,1),(0,0,0,1) \rangle$. If $(r_1 + r_2) \cap r_3 = \emptyset$, then we get that $r \cap r_3 = \emptyset$ which is a contradiction; thus it must follow that $(r_1 + r_2) \cap r_3 \not = \emptyset$. The implicit equations of the hyperplane $r_1+r_2$ are $x_2 - x_3 = 0$. Using the fact that every point of $r_3$ can be expressed as $(0,1,\lambda \cdot a, \lambda)$, we calculate de intersection of the subspaces:
$$ 1 - (\lambda \cdot a) = 0 \iff 1 = \lambda \cdot a \iff a \not = 0, \lambda = 1/a $$
Then, if such a line exists, it must go through the point $P_0 = (0,1,1,1/a)$. However, I can't determine whether if that line exists yet, I just know that $a /= 0$ for sure. How should I proceed now? I thought of calculating the planes  $r_1 + P_0$,$r_2 + P_0$ and intersecting them; as if such a line exists must be coplanar both to $r_1,r_2$ and go through $P_0$. Any idea or approach?
 A: For fixed $a\in\mathbb{R}$, the lines $r_1,r_2,r_3$ can be expressed parametrically as
\begin{align*}
r_1(t)&=\,(t,t,t,t)\\[4pt]
r_2(t)&=\,(t,0,0,1)\\[4pt]
r_3(t)&=\,(0,1,at,t)\\[4pt]
\end{align*}
To see that $r_1\cap r_2={\large{\varnothing}}$ and $r_1\cap r_2={\large{\varnothing}}$, note that any point on either of $r_2,r_3$ must have at least one coordinate equal to $0$ and at least one coordinate equal to $1$, hence can't have all $4$ coordinates equal.

To see that $r_2\cap r_3={\large{\varnothing}}$, note that the second coordinate of any point on $r_2$ is equal to $0$, whereas the second coordinate of any point on $r_3$ is equal to $1$.

Thus the lines $r_1,r_2,r_3$ are pairwise disjoint.

Claim:$\;$There is a line $r_4$ which meets each of $r_1,r_2,r_3$ if and only if $a\not\in\left\{0,{\large{\frac{1}{2}}},1\right\}$, and such a line, when it exists, is unique.

Proof:

Suppose there is a line $r_4$ which meets $r_1,r_2,r_3$  at $P_1,P_2,P_3$, respectively.

Let $t_1,t_2,t_3$ be such that $P_1=r_1(t_1),P_2=r_2(t_2),P_3=r_3(t_3)$.

Then the nonzero vectors $p=P_2-P_1$ and $q=P_3-P_1$ must be parallel, hence each must be a nonzero scalar multiple of the other.

Evaluating $p,q$ we get
\begin{align*}
p&=\,\langle{t_2-t_1,-t_1,-t_1,1-t_1}\rangle\\[4pt]
q&=\,\langle{-t_2,1,at_3,t_3-1}\rangle\\[4pt]
\end{align*}
Looking at the second coordinates of $p,q$, we must have $p=-t_1q$.

It follows that $t_1\ne 0$, and then, equating the other $3$ coordinates of $p$ and $-t_1q$, we get the system
\begin{align*}
t_2-t_1&=t_1t_2&&(\text{eq}1)\\[4pt]
-t_1&=-at_1t_3&&(\text{eq}2)\\[4pt]
1-t_1&=-t_1(1-t_3)&&(\text{eq}3)\\[4pt]
\end{align*}
of $3$ equations in the $3$ unknowns $t_1,t_2,t_3$.

From $(\text{eq}2)$, since $t_1\ne 0$,we get $a\ne 0$ and $t_3={\large{\frac{1}{a}}}$.

Then from $(\text{eq}3)$ we get
$$
a(2t_1-1)=t_1
$$
which implies $a\ne 0$, and we also get
$$
t_1(2a-1)=a
$$
which implies $a\ne {\large{\frac{1}{2}}}$ and $t_1={\Large{\frac{a}{2a-1}}}$.

And then from $(\text{eq}1)$ we get
$$
t_2(a-1)=a
$$
which implies $a\ne 1$ and $t_2={\Large{\frac{a}{a-1}}}$.

Thus we've shown that if there is a line $r_4$ which meets each of $r_1,r_2,r_3$, then $a\not\in\left\{0,{\large{\frac{1}{2}}},1\right\}$, and the points $P_1,P_2,P_3$ where $r_4$ meets $r_1,r_2,r_3$, respectively, are uniquely determined by
$$
P_1=r_1(t_1),\;\;\;P_2=r_2(t_2),\;\;\;P_3=r_3(t_3)
$$
where
$$
t_1=\frac{a}{2a-1},
\;\;\;
t_2=\frac{a}{a-1},
\;\;\;
t_3=\frac{1}{a}
$$
It follows that if such a line $r_4$ exists, it is unique.

To finish the proof of the claim, it suffices to show that for $a\not\in\left\{0,{\large{\frac{1}{2}}},1\right\}$, the points $P_1,P_2,P_3$, defined as above, are collinear, which follows from the easily verified fact that the equation
$$
P_2-P_1=-t_1(P_3-P_1)
$$
is an identity.
