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.
To ensure the sides are integers, we use the famous parametrization:
$$a=m^2-n^2$$
$$b=2mn$$
$$c=m^2+n^2$$
Then we have: ($p$ is perimeter)
$$p=a+b+c=2m^2+2mn=2m(m+n)$$
When $p$ is odd, no integer solutions for $m, n$ exist.
When $p=2ab$, the number of solutions is the number of $b$ such that $a<b<2a$, which is equivalent to the number of $k$ such that $k$ is a factor of $p$ and $\sqrt{p/2}<k<\sqrt{p}$.
For even $p \le 10$, there is no solution.
The first solution occurs at $p=12$, we have $m=2, n=1$, and $(a,b,c)=(3,4,5)$.
EDIT: I realized that a solution rarely exists. If you want to find some $p$ such that a solution exists, I suggest you substitute arbitrary values of $m,n$ into the formula $p=2m(m+n)$.
Best Answer
Hints for part a:
Show that
Hints for part b: