This is simple, but I might have missed some basic case. Anyhow. If $c=2$, then $a=b+4$, and thus $a\equiv b\pmod{4}$. This yields, $a+b+c=2b+6$, and this case will not yield any triple $(a,b,c)$ with $abc<67\cdot 41\cdot 13$, where I define by $P$ the product $67\cdot 41 \cdot 13$. Now, suppose $c$ is odd. Then, $a\equiv b+2\pmod{4}$, and thus, $a+b+c=2(b+1)+c\equiv c\pmod{4}$. Result of this, we get $c\equiv 1\pmod{4}$, thus, $c\geqslant 5$. Similarly, the case $c=5$ does not yield any other solutions. Now, for $c=13$, we get that $a=b+26$, yielding $2b+39$ is a perfect square. Inspecting 'small' cases, we get $b=41$ makes this $11^2$, and makes $a$ a prime. Thus, the smallest (in the sense of product) brilliant pair is $(a,b,c)=(67,41,13)$.

To the signature grypyorum, you are missing a smaller solution than $(a,b,c) = (67,41,13)$ in the case where $c=13$, which means $a=b+26$ is a prime and $2b+39$ is a perfect square. Choosing $b=5$ these two conditions are satisfied, and we obtain $(a,b,c) = (31,5,13)$ as the minimal solution.

Solar Plexsus: We have $a>b>c$.