No, none of the lengths can be prime.
Let me write $d,e,f$ for $h_a,h_b,h_c$, then twice the area is
$$
2\mathrm{Area} = ad=be=cf
$$
Furthermore let $c=x+y$ with $x,y$ possibly signed lengths satisfying $x^2+f^2=a^2$ and $y^2+f^2=b^2$, so $x,y$ are (possibly signed) integers.
Consider the case in which a side length is prime, wlog let it be $c$.
Then since $c>d$, $c\nmid d$, but $c\mid ad$ so we must have $a=tc$, and similarly $c>e$ so $b=uc$ for integers $t,u$. Since $a,b,c$ make a triangle $a<b+c$ and $b<a+c$ we have $t<u+1,u<t+1\implies t=u$ and the triangle must be isosceles with $a=b$.
Then $x=y=c/2$, but $a^2=f^2+x^2$ requires $x$ to be an integer, so $c$ is even and can only be prime if $c=2$. But then $a^2=f^2+1$ which is impossible for positive integers $a,f$, so there cannot be such a triangle with a prime side.
Consider the case in which a height is prime, wlog let it be $f$.
If $f\mid a$ then $f\mid x$ and $(x/f)^2+1=(a/f)^2$ which is impossible for nonzero integers $x/f,a/f$. So $f\nmid a$ and $f\mid ad \implies d=tf$. Similarly $f\nmid b$ and so $e=uf$. Then from $ad=be=cf$ we have $c=ta=ub$. Wlog $a\ge b$, then from $c<a+b$ we have $(t-1)a<b \implies t=1$ and the triangle must be isosceles with $a=c$.
Then with $c=ub,e=uf$ and by symmetry the altitude bisecting $b$
$$
(b/2)^2+e^2=c^2 \\
(b/2)^2+u^2f^2=u^2b^2 \\
f^2=\left(\frac{b}{2u}\right)^2(4u^2-1)
$$
Since $f$ is an integer and $\gcd(2u,4u^2-1)=1$ we must have $2u\mid b$. Since we have already shown $f\nmid b$ and by assumption $f$ is prime, then $\gcd(f,b)=1$ and we must have $b/2u=1$. Then $f^2=4u^2-1$, but this is impossible for positive integers $u,f$, so there cannot be such a triangle with prime height.
Consider two right triangles, with hypotenuses $p$ and $q$ and respective acute angles $\theta$ and $\phi$. To see that having equal perimeter and area makes them congruent, it suffices to show that
$$\theta = \phi \qquad\text{or}\qquad \theta+\phi=\frac{\pi}{2} \tag{0}$$
(Either makes the triangles similar, which in turn makes them congruent.)
Equating perimeters and areas gives a system we can write as
$$\begin{align}
p(1+\sin\theta+\cos\theta) &= q(1+\sin\phi+\cos\phi) \\
p^2 \sin\theta \cos\theta &= q^2 \sin\phi \cos\phi
\end{align} \tag{1}$$
Defining $u:=\tan(\theta/2)$, we "know" that
$$\sin\theta = \frac{2u}{1+u^2} \qquad \cos\theta=\frac{1-u^2}{1+u^2} \quad\to\quad 1+\cos\theta+\sin\theta= \frac{2 (1 + u)}{1 + u^2}$$
and likewise for $v:=\tan(\phi/2)$. Thus, $(1)$ can be rewritten as
$$\begin{align}
p\frac{(1+u)}{1+u^2} &= q\frac{(1+v)}{1+v^2} \\[4pt]
p^2\frac{u(1+u)(1-u)}{(1+u^2)^2} &= q^2\frac{v(1+v)(1-v)}{(1+v^2)^2}
\end{align}\tag{2}$$
Dividing the second equation by the square of the first ...
$$\frac{u(1-u)}{1+u} = \frac{v(1-v)}{1+v} \quad\to\quad (u-v)(uv+u+v-1)=0 \quad\to\quad u=v, \text{ or } \frac{u+v}{1-uv}=1 \tag{3}$$
Therefore, we have one of the following situations (bearing in mind that $\theta/2$ and $\phi/2$ are each at most $\pi/4$, so that we may draw appropriate conclusions from these tangent inequalities):
$$\begin{align}
\tan\frac{\theta}{2}=\tan\frac{\phi}{2} &\quad\to\quad \theta=\phi \\[4pt]
\frac{\tan(\theta/2)+\tan(\phi/2)}{1-\tan(\theta/2)\tan(\phi/2)} = 1 &\quad\to\quad
\tan\left(\frac{\theta}{2}+\frac{\phi}{2}\right)=\tan\frac{\pi}{4} \quad\to\quad \theta+\phi=\frac{\pi}{2}\end{align} \tag{4}$$
which match the sufficient conditions in $(0)$. $\square$
Best Answer
Modulo $6$ the residues of the primes are $3,-1,1,-1,1$ respectively – and our target has residue $-1$ modulo $6$. If we use two $3$s the last side must be $23$, which is impossible. If we use one $3$ the other two sides must have residue $1$ each and sum to $26$ – the only possibility is $13$ and $13$. If no $3$s, there must be two residue $-1$ numbers and one residue $1$ number:
Hence there are $3$ incongruent admissible triangles: $\{3,13,13\},\{7,11,11\},\{5,11,13\}$.