6. Let the lengths of the three sides of a triangle be $a$, $b$, and $c$ satisfying the system of equations
$$
\left\{\begin{array}{l}
2 a b+c^{2}-a=91, \\
2 b c-a^{2}+b=24, \\
2 a c-b^{2}-c=61 .
\end{array}\right.
$$

Then the value of $a+b-c$ is ( ).
(A) -2 or 4
(B) 3
(C) -2 or 3
(D) 4
6. B.

From (1)-(2) - (3), we get
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+2 a b-2 b c-2 a c-a-b+c=6 \text {, } \\
\text { i.e., }(a+b-c)^{2}-(a+b-c)-6=0 \text {. } \\
\end{array}
$$

Solving, we get $a+b-c=3$ or -2.
Since $a, b, c$ are the lengths of the three sides of a triangle, then
$$
a+b-c>0 \text {. }
$$

Therefore, $a+b-c=3$.