Weird Answer involving Minimum

You do not explain how you derive your governing equation; things might have gone wrong somewhere there. Consider this:

If we make the point $A$ our origin of coordinates, then $B$ becomes $(12,0)$ and $C$ becomes $(m,3)$, where $0<m<12,$ indicates the length of $AD.$ The distances from our point $(m,y)$ from the three vertices is as follows: $$\begin{align}{|PA|=\sqrt{y^2+m^2}\\ |PB|=\sqrt{y^2+(m-12)^2}\\ |PC|=\sqrt{(y-3)^2}\\ }\end{align},$$ so that it is their sum that you want to minimize as $y$ varies in the bounds $[0,3].$

You should be able to continue from here?

Edit: When I took note of the fact that $|AC|=|CB|,$ so that consequently, $m=6,$ and simplified, differentiated and set to nullity, I got $$4y^2(y-3)^2=(3-y)^2(y^2+36).$$ It is now easy to see where OP lost his solution, since it is very tempting to cancel the equal factors; but recall that $3$ is in the domain of $y,$ so that we cannot legitimately cancel. Thus, letting $K=(3-y)^2,$ we get $$(4y^2-y^2-36)K=0,$$ from which we also get the possibility $y=3.$

Hope OP doesn't bang his head on a wall. :)

First observe that your function $d$ is a decreasing function of $x (\because$ its derivative is negative $\forall x\in[0,3])$.

Now, since we are restricted in this interval only, the required minimum occurs at $x=3$.

Hope it helps:)