5. Given three pairwise intersecting lines, not parallel to one plane, and a point $P$ not belonging to any of them, construct a plane through this point so that it forms equal angles with the given lines.

保留源文本的换行和格式，这里直接输出翻译结果。
5. Draw through point $P$ the lines $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$, respectively parallel to the given lines $a, b$, and $c$. Point $P$ divides each of the lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$ into two opposite rays $P A$ and $P A^{\prime}, P B$ and $P B^{\prime}, P C$ and $P C^{\prime}$. The planes of symmetry of the rays $P A$ and $P B, P B$ and $P C$ intersect along the line $l$, through which the plane of symmetry of the rays $P C$ and $P A$ also passes (why?). The line $l$ forms congruent angles with the rays $P A, P B$, and $P C$. This same line $l$ forms congruent angles with the opposite rays $P A^{\prime}, P B^{\prime}$, and $P C^{\prime}$. The plane passing through $P$ perpendicular to $l$ forms congruent angles with the lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$, and thus also with the lines $a, b$, and $c$.

Another solution is obtained by considering the rays $P A, P B$, and $P C^{\prime}$ and their opposite rays $P A^{\prime}, P B^{\prime}$, and $P C$. Two more solutions correspond to the triples of rays $P B, P C$, and $P A^{\prime}$ (with their opposites $P B^{\prime}, P C^{\prime}, P A$) and $P A, P C, P B^{\prime}$ (with their opposites $P A^{\prime}, P C^{\prime}$, $P B$). Thus, there are 4 planes that satisfy the condition of the problem.