Let $AB=3$, $AC=4$, $BC=5$. Let $D$, $E$, $F$ be the projections of $P$ into $AB$, $AC$ and $BC$. Assume 1 or 2 or 3 of the distances from $P$ to sides are not rational, and use the fact that $AEPD$ is rectangle. We easily get to a contradiction using Pythagorean theorem .

I used heron's theorem to prove this problem we can calculate space of triangles APC, PCB, PAB suppose n1. r1 and n2. r2 and n3. r3 are respectly the space of triangles and r1 and r2 and r3 are square root of the most little nature number that it can be, and n1, n2, n3 are rational and n1r1 + n2r2+ n3r3=6 now easily we can prove that r1 and r2 and r3 =1 now the space of each triangles is distance between p to sides of triangle multiplied by 3 or 4 or 5 and you can see proof is completed now