We have two vectors, $x = (1,4, -1)$ and $y = (-1,0,1)$ in $\mathbb{R}^3$. We have the following hyperplanes:
$P_1 = \{ v \in \mathbb{R}^3 : \langle x, v \rangle =2\} $
$P_2 = \{ v \in \mathbb{R}^3 : \langle y, v \rangle =0\} $
Show the the intersection of these hyperplanes is a line.
So I think that $P_1$ and $P_2$ are $x_1 + 4x_2 - x_3 = 2$ and $x_1 = x_3$ respectively. So to find the intersection you fill in $x_1 = x_3$ in $P_1$ and you get $x_2=\dfrac{1}{2}$, which is a line.
So now my question:
Is this a satisfactory answer to the question?
What relation does $x$ have with $P_1$ (algebraically or geometrically), or $y$ with $P_2$? It seems as if you're just taking the coordinates of the vector and then multiplying them with the corresponding variable.
I think that it is better to write explicitly the equation of the line in the vector form: $ \vec x=t \vec u + \vec x_0$. In your case, you have found that:
$x_1=x_3=t$ and $x_2=\dfrac{1}{2}$, so we can write the equation: $$ \vec x=t(1,0,1)^T+(0,1/2,0)^T $$