Example 6 If three lines $a, b, c$ in space are pairwise skew lines, then the number of lines that intersect $a, b, c$ is ( ).
(A)0
(B) 1
(C) more than 1 but finite
(D) infinitely many
Analysis: If $a$, $b$, and $c$ are not parallel to the same plane, by Basic Conclusion 8, we can construct a parallelepiped $ABCD-A_{1}B_{1}C_{1}D_{1}$, such that $a$, $b$, and $c$ are its non-parallel edges $AB$, $B_{1}C_{1}$, and $DD_{1}$. Through a point $E$ on $DD_{1}$, different from $D$ and $D_{1}$, and the line $AB$, we can construct a plane $M$. Then, $M$ must intersect the line $B_{1}C_{1}$ at a point $F$. The lines $EF$ and $AB$ are both in plane $M$ and are not parallel, so they must intersect at a point $G$. Thus, a point $E$ on $DD_{1}$, different from $D$ and $D_{1}$, can determine a line that intersects $a$, $b$, and $c$;

If $a$, $b$, and $c$ are all parallel to the same plane, through a point $A$ on $a$ and the line $b$, we can construct a plane $\beta$, then $\beta$ must intersect the line $c$ at a point $C$ (otherwise $c \parallel \beta \Rightarrow a \parallel \beta \Rightarrow a \in \beta$, contradiction). Let $AC$ and $a$ be coplanar in $\gamma$. If $AC \parallel b \Rightarrow b \parallel \gamma \Rightarrow c \parallel \gamma \Rightarrow c \in \gamma$, contradiction. Therefore, $AC$ must intersect $b$ at a point $B$. Thus, a point $A$ on $a$ can determine a line that intersects $a$, $b$, and $c$. Hence, the correct choice is $(\mathrm{D})$.