\begin{cases}
ab+a+b=250 \\
bc+b+c=300 \\
ac+a+c=216 \\
\end{cases}
it seems of no difficulty: a system of 3 equations and 3 variables...
Let's start by adding 1 to every equations, we obtain:
\begin{cases}
ab+a+b+1=(a+1)(b+1)=251 \\
bc+b+c+1=(c+1)(b+1)=301 \\
ac+a+c+1=(a+1)(c+1)=217 \\
\end{cases}
Now let's substitute $a+1 = x, b+1 =y, c+1 = z$, so we can reduce the amount of calculus needed, in fact to solve
\begin{cases}
xy=251 \\
yz=301 \\
xz=217 \\
\end{cases}
you only need to find $x$ (or $y$) in the first equation, substitute it in the last (or second) equation, and then do another substitution. Then subtract 1, and you'll find the solutions:
$(a,b,c) = (-1-\sqrt{\frac{7781}{43}}, -1-\sqrt{\frac{10793}{31}}, -1-7\sqrt{\frac{1333}{251}})$
and
$(a,b,c) = (\sqrt{\frac{7781}{43}}-1, \sqrt{\frac{10793}{31}}-1, 7\sqrt{\frac{1333}{251}}-1)$