11. If real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x-4 y+3 \leqslant 0, \\
3 x+5 y-25 \leqslant 0, \\
x \geqslant 1,
\end{array}\right.
$$

the maximum value of the objective function $z=k x+y$ is 12, and the minimum value is 3, then the value of the real number $k$ is ( ).
(A) 2
(B) -2
(C) $\frac{1}{5}$
(D) does not exist
11. A.

From the system of linear equations, it is easy to find that the intersection points of the three lines are $A\left(1, \frac{22}{5}\right), B(1,1), C(5,2)$.

Substituting $k=2$ into the objective function, the intercept $z=3$ through point $B$, and the intercept $z=12$ through point $C$, which meets the requirements of the problem. Therefore, $k=2$.