3. Three positive integers $a$, $b$, and $c$ satisfy the conditions:
(1) $a<b<c<30$;
(2) For some positive integer base, the logarithms of $a(2 b-a)$ and $c^{2}+$ $60 b-11 a$ are 9 and 11, respectively.

Then the value of $a-2 b+c$ is $\qquad$.
3. -4 .
Let the base be a positive integer $n(n \geqslant 2)$, then from condition (2) we get
$$
\left\{\begin{array}{l}
a(2 b-a)=n^{9} . \\
c^{2}+60 b-11 a=n^{11} .
\end{array}\right.
$$

When $n \geqslant 3$, from equation (1) we get
$$
3^{9} \leqslant n^{9}=a(2 b-a)2^{9}$, i.e., $b \geqslant 23$.
$$
Thus, we have
$$
\left\{\begin{array}{l}
a(2 b-a)=2^{9}, \\
c^{2}+60 b-11 a=2^{11} .
\end{array}\right.
$$

Combining that $a, b, c$ are positive integers, and $a<b<c<30, b \geqslant 23$, we get $a=16, b=24, c=28$.
Therefore, $a-2 b+c=-4$.