Step 1 :The function \(f(x)=a x^{2}+11\) is a quadratic function, and the graph of a quadratic function is symmetric about the line \(x=h\), where \(h\) is the x-coordinate of the vertex.
Step 2 :In this case, the vertex is the minimum point of the function, and since the coefficient of \(x^2\) is positive, the graph opens upwards.
Step 3 :Therefore, the vertex is the minimum point of the function, and the x-coordinate of the vertex is \(-b/2a\), where \(b\) is the coefficient of \(x\) and \(a\) is the coefficient of \(x^2\).
Step 4 :Since there is no \(x\) term, the x-coordinate of the vertex is 0. Therefore, the function is symmetric about the line \(x=0\), which means that \(f(6)=f(-6)\).
Step 5 :Given that \(f(6)=27\), we can find that \(a = 0.4444444444444444\).
Step 6 :Substituting \(a\) into the function, we can find that \(f(-6) = f(6) = 27\).
Step 7 :So, the correct answer is \(\boxed{27}\).