The parabola $y = x^2 + bx + c$ has the following properties:

The point on the parabola closest to $(12,3)$ is the $y$-intercept of the parabola.
The parabola passes through $(-5,0).$

Enter the ordered pair $(b,c).$
The $y$-intercept is $(0,c).$  Since this is the point closest to $(12,3),$ the line joining $(0,c)$ and $(12,3)$ is perpendicular to the tangent to the parabola at $(0,c).$

[asy]
unitsize(0.5 cm);

real parab (real x) {
  return(x^2 + 6*x + 5);
}

draw(graph(parab,-6.5,0.5),red);
draw((-7,0)--(15,0));
draw((0,-5)--(0,10));
draw(((0,5) + (5)*(1/6,1))--((0,5) + (-8)*(1/6,1)),dashed);
draw((0,5)--(12,3));

dot("$(12,3)$", (12,3), E);
dot("$(-5,0)$", (-5,0), SW);
dot("$(0,c)$", (0,5), W);
[/asy]

The equation of the tangent is of the form
\[y - c = mx\]for some real number $m,$ so $y = mx + c.$  Substituting into $y = x^2 + bx + c,$ we get
\[mx + c = x^2 + bx + c,\]so $x^2 + (b - m) x = 0.$  Since $y = mx + c$ is the equation of the tangent at $x = 0,$ this quadratic should have a double root of $x = 0,$ which means $m = b.$

Thus, the slope of the tangent is $b.$  The slope of the line joining $(0,c)$ and $(12,3)$ is $\frac{3 - c}{12},$ so
\[b \cdot \frac{3 - c}{12} = -1.\]Then $b = -\frac{12}{3 - c} = \frac{12}{c - 3}.$

Also, the parabola passes through $(-5,0),$ so
\[0 = 25 - 5b + c.\]Substituting $b = \frac{12}{c - 3},$ we get
\[25 - \frac{60}{c - 3} + c = 0.\]This simplifies to $c^2 + 22c - 135 = 0,$ which factors as $(c - 5)(c + 27) = 0.$  Hence, $c = 5$ or $c = -27.$

If $c = -27,$ then $b = -\frac{2}{5},$ which does not satisfy the given conditions.  Therefore, $c = 5,$ and $b = 6,$ so $(b,c) = \boxed{(6,5)}.$