jmc

$\text{Solution 1}$: Let the three numbers be $a$, $b$, and $c$, and WLOG assume that $a\le b\le c$. We have the three equations $$\begin{array}{rl}a+b+c& =67\\ c-b& =7\\ b-a& =3\end{array}$$ From the second equation, we have $c=b+7$. Substituting this into the first equation to eliminate $c$, we have $a+b+(b+7)=67\Rightarrow a+2b=60$. Adding this last equation to the third equation, we have $a+2b+b-a=60+3\Rightarrow b=21$. Substituting this value into the second equation to find $c$, we get $c=b+7=28$. Thus, the largest number is $\boxed{28}$.

$\text{Solution 2}$: Let the middle number be $x.$ Then, the largest number is $x+7$ and the smallest number is $x-3.$ The numbers have a sum of $67,$ so we have the equation $$(x-3)+(x)+(x+7)=67.$$ Simplifying, we get $$3x+4=67$$ $$⟹x=21.$$ So, the largest number is $x+7=21+7=\boxed{28}.$