Equation of line gives equation of plane Given a 3D line in parametric form $$x = 5 + t$$$$y = 1 +3t$$$$z = 4t$$
I did the following calculation: $$x + y + z = (5 + t) + (1 + 3t) + (4t)$$
Therefore       $$x + y + z = 6 + 8t = 6 + 2(4t) = 6 + 2z$$
Finally, $$x + y - z = 6$$
But this is the equation of a plane!!
So where did I go wrong??

A line in $\mathbb{R}^3$ is a subset of $\mathbb{R}^3$. Your set of equations can be written as the set $\{(5+t,1+3t,4t) \in \mathbb{R}^3 : t \in \mathbb{R}\}.$ But when you write $x+y+z$, you've no longer got a set of points in $\mathbb{R}^3$, you've got a set of points in $\mathbb{R}$.
