\[
f(x)=a x^{2}+11
\]
For the function $f$ defined above, $a$ is a constant and $f(6)=27$. What is the value of $f(-6)$ ?
A) 27
B) $\frac{4}{9}$
C) $-\frac{4}{9}$
D) -27

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 \(\boxed{27}\).