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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 $(0,c).$



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 $(b,c)=\boxed{(6,5)}.$