Finding point through a parametric equation

Question

The question is the parametric equations for a line $L_1$ are as follows: $x = 5 - 4t$, $y = 2 - 6t$, $z = -1-2t$.

So I'm guessing another way to format that line would be $(5 -4t, 2 -6t, -1-2t)$.

Let $L_2$ be the line parallel to $L_1$ and passing through the point $(1,-2,3)$. Find the point $P$ on $L_2$ whose $x$ coordinate is $-9$.

I'm confused on what to do.

Answer 1

HINT

Line $L_2$ is

$$(1,-2,3)+t(-4,-6,-2)=(1-4t,-2-6t,3-2t)$$

note also that

$$x=1-4t=-9\implies t=\frac52$$