13. Given integers $a$, $b$, and $c$ such that the equation
$$
(x+a)(x+b)+c(x-10)=(x-11)(x+1)
$$

holds for all $x$. Find the value of $c$.
Three, 13. Expanding the equation in the question, we get
$$
x^{2}+(a+b+c) x+a b-10 c=x^{2}-10 x-11 \text {. }
$$

The above equation holds for any $x$. Therefore,
$$
\left\{\begin{array}{l}
a+b+c=-10, \\
a b-10 c=-11 .
\end{array}\right.
$$

Eliminating the parameter $c$, we get
$$
10 a+10 b+a b=-111 \text {, }
$$

which is $\square$
$$
(a+10)(b+10)=-11 \text {. }
$$

Since $a$ and $b$ are integers, and
$$
-11=(-1) \times 11=1 \times(-11) \text {, }
$$

we have, $\left\{\begin{array}{l}a+10=1,-1,11,-11, \\ b+10=-11,11,-1,1 .\end{array}\right.$
Therefore, $a+b+20=-10,10$, which means
$$
a+b=-30,-10 \text {. }
$$

Thus, $c=20$ or 0.