11. Given the quadratic function $y=x^{2}+2 m x-3 m+1$, the independent variable $x$ and real numbers $p, q$ satisfy
$$
4 p^{2}+9 q^{2}=2, \frac{1}{2} x+3 p q=1 \text {, }
$$

and the minimum value of $y$ is 1. Find the value of $m$.
Three, 11. From the given information,
$$
\begin{array}{l}
(2 p+3 q)^{2}=2+12 p q, 6 p q=2-x \\
\Rightarrow 2 p \times 3 q=2-x, 2 p+3 q= \pm \sqrt{6-2 x} .
\end{array}
$$

Then $2 p, 3 q$ are the two real roots of the equation in $t$:
$$
t^{2} \mp \sqrt{6-2 x}+2-x=0
$$

Thus, $\Delta=(6-2 x)-4(2-x)=2 x-2 \geqslant 0$, which means $x \geqslant 1$.
Also, $6-2 x \geqslant 0$, so $1 \leqslant x \leqslant 3$.
It is easy to see that the vertex of the quadratic function is
$$
\left(-m,-m^{2}-3 m+1\right) \text {. }
$$
(1) If $m \leqslant-3$, when $x=3$, $y$ is the smallest, i.e.,
$$
9+6 m-3 m+1=1 \Rightarrow m=-3 \text {; }
$$
(2) If $-3<m<-1$, then
$$
-m^{2}-3 m+1=1 \text {, }
$$

no solution;
(3) If $m \geqslant-1$, then
$$
1+2 m-3 m+1=1 \Rightarrow m=1 \text {. }
$$

In summary, $m=-3$ or 1.