11. (20 points) If the equation about $x$
$$
x^{2}+k x-12=0 \text { and } 3 x^{2}-8 x-3 k=0
$$

have a common root, find all possible values of the real number $k$.
Three, 11. Let $x=x_{0}$ be the common root given in the problem.
Then $\left\{\begin{array}{l}x_{0}^{2}+k x_{0}-12=0, \\ 3 x_{0}^{2}-8 x_{0}-3 k=0 .\end{array}\right.$
(1) $\times 3$ - (2) gives $(3 k+8) x_{0}=36-3 k$.

Obviously, $k \neq-\frac{8}{3}$.
Thus, $x_{0}=\frac{36-3 k}{3 k+8}$.
Substituting equation (3) into equation (1) yields
$$
\begin{array}{l}
\left(\frac{36-3 k}{3 k+8}\right)^{2}+k \cdot \frac{36-3 k}{3 k+8}-12=0 \\
\Rightarrow 9 k^{3}+15 k^{2}+504 k-528=0 \\
\Rightarrow(k-1)\left(9 k^{2}+24 k+528\right)=0 .
\end{array}
$$

Since the discriminant of $9 k^{2}+24 k+528=0$ is
$$
\Delta=24^{2}-4 \times 9 \times 5280$.
Thus, $k=1$.
Upon verification, when $k=1$, $x_{0}=3$ satisfies the conditions.