equation of 3d line from system of equation

Question

The line is the intersection between the planes

$x+3y=7$ and $2y+z=4$

from this I get the line on the form

$$x = 7-3\frac{4-t}{2}$$ $$y = \frac{4-t}{4}$$

How did they in my textbook go from this form to the vector form of the line:

$(x, y, z) = (1, 2, 0)+t(3, −1, 2)$.

Answer 1

You have $t=z$ and your equations give

$x=1+\frac{3}{2}t, y=2-\frac{1}{2}t, z=t$, with $s=2t$ we derive

$$(x, y, z) = (1, 2, 0)+s(3, −1, 2).$$

Answer 2

They found a point belonging to both planes: $A=(1,2,0)$ and used the cross-product of the normal vectors $\vec n_1 (1,3,0)$ and $\vec n_2(0,2,1)$ as the directing vector of the line. The parametric equation then becomes: $$M=A+t\mkern1mu(\vec n_1\times \vec n_2).$$