Step 1: Write the given conditions.
We are given
\[
f(1)+2f(2)=0
\]
and
\[
2f(1)+f(2)=0
\]
Let
\[
f(1)=p
\]
and
\[
f(2)=q
\]
Then the equations become
\[
p+2q=0
\]
and
\[
2p+q=0
\]
Step 2: Solve for \(p\) and \(q\).
From
\[
p+2q=0,
\]
we get
\[
p=-2q
\]
Substitute this in
\[
2p+q=0
\]
Then,
\[
2(-2q)+q=0
\]
\[
-4q+q=0
\]
\[
-3q=0
\]
\[
q=0
\]
Therefore,
\[
p=0
\]
Hence,
\[
f(1)=0
\]
and
\[
f(2)=0
\]
Step 3: Use the quadratic polynomial.
Given,
\[
f(x)=ax^2+bx+c
\]
Now,
\[
f(1)=a+b+c=0
\]
Also,
\[
f(2)=4a+2b+c=0
\]
Step 4: Subtract the equations.
Subtract
\[
a+b+c=0
\]
from
\[
4a+2b+c=0
\]
We get
\[
(4a+2b+c)-(a+b+c)=0
\]
\[
3a+b=0
\]
Step 5: Final conclusion.
Therefore,
\[
\boxed{0}
\]