We have the equations $g_1: (4,14,-5)+t(2,-3,0)$ and $g_2:(11,9,-15)+t(-2,-3,2)$ on which two boats move at time $t$. We have the point $P(12,11.5,0)$.
Is it possible that the one boat is behind the other one when we look from the point $P$?
Do we have to find the equation of the line between $P$ and $(2,-3,0)$ and check if there is a t such that $g_2$ is on that line?
On
$$ \exists\;t |\; \vec{g}_1+\lambda(\vec{g}_2 - \vec{g}_1) = \vec{OP\;} $$ $$ \vec{g}_2 - \vec{g}_1 = (11-4,9-14,-15+5)+t(-2-2,-3+3,2-0) =(7,-5,-10)+t(-4,0,2) $$ $$ \vec{g}_1+\lambda(\vec{g}_2 - \vec{g}_1) = (4,14,-5)+t(2,-3,0)+\lambda((7,-5,-10)+t(-4,0,2)) = (4+7\lambda-4t\lambda+2t,14-3t-5\lambda,-5-10\lambda+2t\lambda)=(12,11.5,0) $$
You don't need to do this for the point $(2,-3,0)$. Rather you need to consider three points $P, P_1(t), P_2(t)$ and find whether there is a value of $t$ for which these three lie on a straight line. $P_1, P_2$ lie on $g_1, g_2$ respectively and are the points associated with time $t$.
There are two procedures you might consider. The first is to compute the line joining $P_1(t)$ and $P_2(t)$ and then test the condition for $P$ to lie on that line.
The second would be to find the point $Q_2$ at which the line joining $P$ and $P_1(t)$ meets $g_2$ - this will be associated with a time $s$. Then test the condition for $s=t$.