The sought line $t$ is either 1) incident to the plane $[a b]$ or 2) the intersection point $(a b)=K$. Apart from these two cases, which do not exclude each other, no other case is possible.

1) Let the intersection points of the lines $c$ and $d$ with the plane $[a b]$ be $P$ and $Q$, respectively. The line $P Q=t_{1}$ is clearly one solution.
2) The intersection point $(a b)=K$ with the lines $c$ and $d$ determines the planes $[K c]$ and $[K d]$, respectively. The intersection line of these two planes $|[K c][K d]|=t_{2}$ is the other solution.

Therefore, there are always 2 and only 2 solutions.

If the line $P Q$ is incident to the point $K$ (or, equivalently, the intersection line of the planes $[K c]$ and $[K d]$ lies in the plane $[a b]$), then the two solutions coincide.

If the plane $[a b]$ is parallel to $c$ (or $d$), then $P$ (or $Q$) goes to infinity and $t_{1} \| c$ (or $d$).

If the plane $[a b]$ is parallel to both $c$ and $d$, then the line at infinity in the plane $[a b]$ is the sought $t_{1}$.
In space, let $a$ and $b$ be intersecting lines, and $c$ and $d$ be any lines that are skew to each other and also skew to both $a$ and $b$. Determine a line that intersects all four lines.