Finding vector equation of a plane from its Cartesian equation The Cartesian equation is $x-3y-4z=1$. Here is what I have tried:
Finding three points on the plane by setting two variables equal to 0:
$x=0$, $y=0$; $z=\frac{-1}{4}$
$y=0$, $z=0$; $x=1$
$x=0$, $z=0$; $y=\frac{-1}{3}$  
Vector equation: $(x,y,z)=(1,0,0)+\lambda[(0,0,\frac{-1}{4})-(0,\frac{-1}{3},0)] + t((0,0,\frac{-1}{4})-(1,0,0)]$
$(x,y,z)=(1,0,0)+\lambda(0,\frac{1}{3},\frac{-1}{4})+t(-1,0,\frac{-1}{4})$  
However, the solution gives the vector equation as: $(x,y,z)=(1,0,0)+\lambda(3,1,0)+t(4,0,1)$. I understand that there are multiple ways to find the vector equation of a plane. But since I am doing this for transformation purposes, the vector equation I found is a little more complicated than the solution's equation. How can I derive the solution's vector equation?
I would appreciate if someone can assist me with this. Thanks
 A: I'll show you how to do a similar one (yours goes exactly the same way), starting with the plane $$3x+4y+5z=6.$$ First, one of the variables has a non-zero coefficient. Pick one, I'll choose $z$. Now I'll solve for $z$. $$z = \frac{1}{5}(6-3x-4y)$$ So a given point on our plane looks like 
$$\begin{align*}
(x,y,z) &= \left(x,y,\frac{1}{5}(6-3x-4y)\right) \\
&= \frac{1}{5}\left(5x,5y,6-3x-4y\right) \\
&= \frac{1}{5}\left[(5x,0,-3x)+(0,5y,-4y)+(0,0,6)\right]\\
&=\frac{x}{5}(5,0,-3)+\frac{y}{5}(0,5,-4)+\left(0,0,\frac{6}{5}\right)
\end{align*}$$
Now, to clean things up, I'll change variables to $s=x/5$ and $t=y/5$ to get $$s(5,0,-3)+t(0,5,-4)+\left(0,0,\frac{6}{5}\right)$$ as a parametrization of the plane.
A: Let be $P_0=\left(0,0,-\frac{1}{4}\right)$, $P_1=\left(1,0,0\right)$, $P_2=\left(0,-\frac{1}{3},0\right)$.
Two vectors in the plane are
$$\pmb v=4\overrightarrow{P_0P_1}=(P_1-P_0)=\left(4,0,1\right)$$ and $$\pmb u=3\overrightarrow{P_1P_2}=3(P_2-P_1)=\left(3,1,0\right)$$
where we we multiply by 4 and 3 to eliminate the fractions.
So you have
$$
(x,y,z)=\left(1,0,0\right)+\lambda \left(3,1,0\right) +t \left(4,0,1\right)
$$
A: The equation in the given solution might have been derived by writing the homogeneous equation $x-3y-4z=0$ in matrix form: $$\pmatrix{1&-3&-4}\pmatrix{x\\y\\z}=0.$$ The coefficient matrix is already trivially in row-reduced echelon form, so a solution to this equation—the nullspace of the coefficient matrix—can be read directly from it: $\lambda(3,1,0)^T+t(4,0,1)^T$. The solution to the original inhomogeneous equation is then the sum of any particular solution to that equation and this solution to the homogeneous equation. $(x,y,z)^T=(1,0,0)$ is an easy choice for a particular solution, resulting in the equation $$(x,y,z)^T=(1,0,0)^T+\lambda(3,1,0)^T+t(4,0,1)^T$$ for the plane.
