If a right triangle is comprised entirely of integral sides, then the two legs and hypothenuse can be represented by: $a = (p^2-q^2)N; b = 2pqN, c = (p^2+q^2)N$ (for $p,q,N \in \mathbb{N}$ and $p>q$); where $N=1$ is the case of a primitive Pythagorean triple. If the difference between the hypothenuse and one of the legs is $75$, then $N$ must be odd since this same hypothenuse/leg difference would be even-valued for even $N$. This will require (1) an odd $N$, (2) an odd hypothenuse, (3) an even leg: $(p^2+q^2)N - 2pqN = 75 \Rightarrow N(p^2-2pq+q^2) = 75 \Rightarrow N(p-q)^2 =3^{1}5^{2}\Rightarrow N=3, p-q=5$. Starting with $p=6,q=1$ we obtain the triplet: $a = 3(6^2-1^2); b = 3(2 \cdot6 \cdot 1); c = 3(6^2+1^2) \Rightarrow (36, 105, 111) \Rightarrow 111-36= 75$ which is satisfied! Hence, the smallest perimeter is $36+105+111=$$\fbox{252}$.