Find three consecutive positive integers such that the square of their sum exceeds the sum of their squares by 214.

Let the integers be $x-1$, $x$, and $x+1$. Their sum is $3x$, so the condition gives $(3x)^2=(x-1)^2+x^2+(x+1)^2+214$. Expanding, $9x^2=3x^2+216$, hence $6x^2=216$ and $x^2=36$. Since the integers are positive, $x=6$, so the numbers are $5,6,7$.