Let $O$ be the centre of the incircle. Join $O$ to the three vertices.
It is not absolutely clear what ratio $12:5$ means. Let $M$ be the middle of the base. Maybe (i) $BO:OM=12:5$, or maybe (ii) $BO:OM=5:12$. (It will turn out that (ii) can't happen.)
We examine the consequences of (i). Let the base be $2y$ (I don't like fractions).
Let $BO=12t$ and let $OM=5t$.
We calculate the area of triangle $BAC$ in two different ways. The area is half of base times height, so it is $(1/2)(2y)(17t)$, that is, $17yt$.
The area is also the sum of the areas of $\triangle BOA$, $\triangle BOC$, and $\triangle OAC$. This sum is $(1/2)(30)(5t)+(1/2)(30)(5t)+(1/2)(2y)(5t)$, which is $150t +5yt$.
We conclude that
$$17yt=150t+5yt.$$
Thus $12y=150$, and therefore $2y=25$.
Next we examine possibility (ii). The analysis is similar. We find pretty quickly that $y$ is "too big" (the Triangle Inequality is violated).
Let a triangle be inscribed in a unit circle, and let $A$ and $B$ mark two vertices. Let $\theta$ be half the length of the arc connecting $A$ and $B$, and let $\ell$ be the length of the chord, you have that from elementary trigonometry
$$ \ell / 2 = \sin (\theta) $$
Now let $\theta_1, \theta_2, \theta_3$ be half the lengths of the three arcs demarcated by the vertices of the triangle, from the above we have that the perimeter is equal to
$$ 2 \left[ \sin (\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] $$
and we know that
$$ \theta_1 + \theta_2 + \theta_3 = \pi $$
while
$$ \theta_1, \theta_2, \theta_3 \in [0,\pi] $$
Now from the fact that $\sin$ is a concave function on $[0,\pi]$, we have that the perimeter is equal to
$$ 6 * \frac13 \left[ \sin(\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] \leq 6 * \sin (\pi / 3) $$
by Jensen's inequality, with the optimum (being the case of the equality) holding when $\theta_1 = \theta_2 = \theta_3$.
Best Answer
Let us draw a picture for the story:
Here $AB$ is the base with the mid point $O$ (placed in the origin of the axes in the picture), and $C$ is on the side bisector of $AB$. The triangle from the $OP$ is $\Delta ABC$, and we know that it has the incenter $I$ delimiting the proportion $5:12$ on the height $CO$. First of all, the "samller part" is $IO$, because $O$ is the biggest angle in $\Delta CAO$. So we have explicitly, also using the angle bisector theorem in this triangle: $$ \frac{5}{12} = \frac{IO}{IC} = \frac{AO}{AC} = \frac{AB/2}{AC} = \frac{25}{AC} \ . $$ This gives $AC=60$.
$\square$
The picture was drawn using scale one to five.