I'm currently studying Analytic Geometry in $\mathbb{R}^3$ and I'm stuck in this exercise:
Find the equation of the line that passes through the point $A\left(\, 1,1,1\, \right)$, that is also parallel to the plane $\pi = x - y + z - 3 = 0$ and intersects the lines $x = 1$ and $y = 3$.
I know that the equation of the line I need to find needs to be perpendicular to the normal vector of the plane, however I'm lost on how I'd also make the line equation intersect those lines $x =1$ and $y = 3$.
The normal vector to your plane is $\langle 1, -1, 1\rangle$. As Carser pointed out, the planes $x = 1$ and $y = 3$ intersect on a line which has the form $\langle 1, 3, z \rangle$. The slope of your line segment will thus be:
\begin{align} \vec{A} - \langle 1, 3, z \rangle =&\ \langle1 - 1, 1 - 3, 1 - z \rangle \\ =&\ \langle 0, -2, 1 - z \rangle \end{align}
This should be perpendicular to the normal vector of the plane thus:
$$ \langle 1, -1, 1\rangle \circ \langle 0, -2, 1 - z \rangle = 0 $$
This gives:
$$ 0 + 2 + 1 - z = 0 \rightarrow 3-z = 0 \rightarrow z = 3 $$
Therefore your line passes through the line segment from $\langle 1,1,1\rangle$ to $\langle 1, 3, 3 \rangle$.