If $x$, $y$, and $z$ are real numbers such that

$(x-3)^2 + (y-4)^2 + (z-5)^2 = 0$,
then $x + y + z =$
$\mathrm{(A)\ } -12 \qquad \mathrm{(B) \ }0 \qquad \mathrm{(C) \  } 8 \qquad \mathrm{(D) \  } 12 \qquad \mathrm{(E) \  }50$
If the sum of three squared expressions is zero, then each expression itself must be zero, since $a^2 \ge 0$ with the equality iff $a=0$.
In this case, $x-3=0$, $y-4=0$, and $z-5=0$.  Adding the three equations and moving the constant to the right gives $x + y + z = 12$, and the answer is $\boxed{D}$.