Check if the line has a $Z$-intercept 
Check if the line $l = ( 3,1,1) + t (1,-1,0)$ has a $Z$-intercept.

Attempt at solution:
Line 1 (Z- intercept): $x = 0, y = 0, z = t$
Line 2 ($l$) = $x = 3+t,y = 1-t,z=1$
Solving this system of equations:
Solving y and z
$$t = 1, t = 1$$
Subbing in to x equation$$0 = 3 + 1, 0 = 4$$
The statement is not true so there is no $Z$-intercept? Not sure if I used the correct process to solve this question.
 A: If the line  $l(t) = ( 3,1,1) + t (1,-1,0)$ were to intersect the $z$-axis, then 
the two equations 
$$
3+t=0 \mbox{ and }1-t=0
$$
must be satisfied, which are obtained by setting the $x$-component and the $y$-component of $l(t)$ equal to $0$. 
This means that $t=-3$ and $t=1$, which is not possible. 
So the line does not intersect the $z$-axis. 
A: This line lives in the plane $z=1$. Changing the value of $t$ will not change the $z=1$ value.
The curve is 
$$
p(t) = 
\left[ \begin{array}{c}
  x(t) \\ y(t) \\ z(t)
\end{array} \right]
=
\left[ \begin{array}{c}
  1+ t \\ 1 -t \\ 1
\end{array} \right]
$$
This is the line $$y=-x+4$$ embedded at $z=1$. This curve never touches the $z=0$ plane. There is no $z-$intercept.
A: It turns out it is true that the line $\ell$ has no $z$ intercept, but your method is not fool-proof. The problem is, you are parametrizing both $\ell$ and the $z$-axis with the same parameter $t$. To see why this is a problem, consider these two 2D curves in the $xy$-plane:
\begin{align}
y &= x \\
y &= 5
\end{align}
These clearly intersect at the point $(5,5)$, but if we parametrize these two curves in just the wrong way, we can make it so it looks like they don't:
\begin{align}
t &\mapsto (2t, 2t) \\
t &\mapsto (t, 5)
\end{align}
If you work thru it, you'll see that there is no $t$ value for which $(2t,2t) = (t,5)$ because you'd get that $t = 0 \textrm{ AND } 5/2$ which is a contradiction.
The reason is best explained if you think of a parametrization of a curve as being a description of how a single point moves across space over time ($t$). $t \mapsto (2t,2t)$ and $t \mapsto(t, 5)$ both trace out the curves $y = x$ and $y = 5$ respectively, but in such a way that the two particles never "collide". To put it another way, there is no $t$ where both points are at the intersection point of $y=x$ and $y=5$ at the same time $t$.
You can see how this affects your 3D case. The curve $\ell$ has a fixed $z$ coordinate of $z=1$, but your parametrization of the $z$-axis moves along the $z$-axis so that the only time it is ever at the point $(0,0,1)$ is at $t=1$. But there's no reason why $\ell(t)$ needs to intersect the $z$-axis at $t=1$. Maybe it decides to intersect at $t=2$. In that case, $\ell(t)$ would cross the path traced out by $z(t)$ (and hence have a $z$-intercept), but $z(t)$, so to speak, wouldn't know it because it has already moved on to the point $(0,0,2)$.
For your particular problem, I think the easiest thing to do is to not create another space curve for the $z$-axis and just work with the parametrization of $\ell$ that you already have. To have a $z$-intercept means that $x = y = 0$ for some $t$ value (we don't care what the $z$ value is, because all $z$ values are on the $z$-axis). So all you really need to do is see if there is a single $t$ value such that $x(t) = y(t) = 0$. Or, in your case,
\begin{align}
3+t &= 0 \\
1-t &= 0
\end{align}
You can see that if both equations were true, it would mean that $t = -3 \textrm{ AND } 1$ which is impossible. Hence there is no such $t$, and $\ell$ has no $z$-intercept.
