6. If in space three lines $a, b, c$ 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
(D)
6. 【Analysis and Solution】Solution 1 First, regardless of the positional relationship between $a$, $b$, and $c$, a parallelepiped $A B C D-A_{1} B_{1} C_{1} D_{1}$ can always be constructed such that $A B$ lies on line $a$, $B_{1} C_{1}$ lies on line $b$, and $D_{1} D$ lies on line $c$. Then, take any point $M$ on the extension of $D D_{1}$. Determine a plane $\alpha$ by $M$ and $a$, and let plane $\alpha$ intersect line $B_{1} C_{1}$ at $P$ and line $A_{1} D_{1}$ at $Q$, then $P Q / / a$. Thus, within plane $\alpha$, line $P M$ does not parallel $\alpha$, and $P M$ must intersect $\alpha$ at a point $N$. Therefore, line $M N$ intersects lines $a$, $b$, and $c$ simultaneously. Since there are infinitely many ways to choose $M$, there are infinitely many lines that intersect $a$, $b$, and $c$ simultaneously. As shown in Figure 2.

Solution 2 Take $a$, $b$, and $c$ as three edges of a cube that are pairwise skew, specifically $A D$, $C C_{1}$, and $A_{1} B_{1}$. Take any point $M$ on $A D$ and point $N$ on $B C$ such that $B_{1} N / / A_{1} M$. Let the line $B_{1} N$ intersect $C C_{1}$ at point $P$. Then $P M$ intersects $a$, $b$, and $c$. Since $M$ is chosen arbitrarily, there are infinitely many lines that satisfy the condition.