We can do this by setting $z=0$ in both plane equations and solving for $x$ and $y$.
From the first equation, $x+2z=12$, we get $x=12$ when $z=0$.
From the second equation, $y-3z=6$, we get $y=6$ when $z=0$.
So, the point $(12,6,0)$ lies on the line of
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