3. Vector Set
$$
\begin{array}{l}
M=\{\boldsymbol{a} \mid \boldsymbol{a}=(-1,1)+x(1,2), x \in \mathbf{R}\}, \\
N=\{\boldsymbol{a} \mid \boldsymbol{a}=(1,-2)+x(2,3), x \in \mathbf{R}\} .
\end{array}
$$

Then $M \cap N=(\quad$.
(A) $\{(1,-2)\}$
(B) $\{(-13,-23)\}$
(C) $\{(-1,1)\}$
(D) $\{(-23,-13)\}$
3. B.

If $\boldsymbol{a}=\left(a_{1}, a_{2}\right) \in M \cap N$, then we have
$$
\begin{array}{l}
a_{1}=-1+x_{1}=1+2 x_{2}, \\
a_{2}=1+2 x_{1}=-2+3 x_{2} .
\end{array}
$$

Rearranging, we get
$$
\left\{\begin{array}{l}
x_{1}-2 x_{2}=2, \\
2 x_{1}-3 x_{2}=-3
\end{array}\right.
$$

Solving, we get $x_{1}=-12, x_{2}=-7$. Therefore, we have
$$
M \cap N=\{(-1-12,1-24)\}=\{(-13,-23)\} \text {. }
$$