For real numbers $x, y$, the algebraic expression
$$
2 x^{2}+3 y^{2}-4 x y-5 x-6 y+7
$$

has the minimum value of ( ).
(A) $-28 \frac{7}{8}$
(B) $-27 \frac{5}{8}$
(C) $-26 \frac{3}{8}$
(D) $-25 \frac{1}{8}$
- 1.C.
$$
\begin{array}{l}
2 x^{2}+3 y^{2}-4 x y-5 x-6 y+7 \\
=2 x^{2}-(4 y+5) x+3 y^{2}-6 y+7 \\
=2\left(x-\frac{4 y+5}{4}\right)^{2}+\left(y-\frac{11}{2}\right)^{2}-26 \frac{3}{8} .
\end{array}
$$

When $x-\frac{4 y+5}{4}=0, y-\frac{11}{2}=0$, i.e., $x=\frac{27}{4}, y=\frac{11}{2}$,
the expression (1) achieves its minimum value of $-26 \frac{3}{8}$.