Let us consider two planes with equations:
$P_1 : x + 2y + 3z = 3$ and $P_2 : 2x-y = 5$
By substituting $x$ as $0$, can I say that the point $(0,-5,13/3)$ lies on the line of intersection of two planes?
And if I find that the direction vector of the line of intersection is $(3,5,-5)$, Can I say that the equation of the line of intersection of two planes is $(0,-5,13/3) + t(3,5,-5)$ ?
I think in the vector $(3,5,-5)$ there is a mistake.
I like the following way.
Let $x=t$.
Thus, $$y=-5+2t,$$ $$z=\frac{13}{3}-\frac{5}{3}t,$$ which gives the answer $$\left(0,-5,\frac{13}{3}\right)+t(3,6,-5).$$
I got it by the following reasoning: $$x=0+1\cdot t,$$ $$y=-5+2t$$ and $$z=\frac{13}{3}-\frac{5}{3}t$$ or since $t$ is any real number, $$x=0+3t,$$ $$y=-5+6t$$ and $$z=\frac{13}{3}-5t$$