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??

Your work is fine.  You found the equation of a plane in which the line lies.  However, not every point in the plane will lie on the line.  That is, there exist $x,y,z$ such that $x+y-z=6$ but that do not satisfy your original parametric equations for any $t\in\mathbb{R}$.  
