$P(a,b,c)=P(bc,ca,ab)$ weighted projective planes for pairwise coprime $a,b,c$

Question

Let $a,b,c\geq 2$ be pairwise coprime integers. The (complex) weighted projective plane $P(a,b,c)$ is the quotient of $\Bbb C^3-\{0\}$ by the action of $\Bbb C^*=\Bbb C-\{0\}$ given by $t\cdot (x,y,z)=(t^ax,t^by,t^cz)$. Or equivalently, the quotient of $S^5\subset \Bbb C^3$ by the $S^1$-action given by $t\cdot (x,y,z)=(t^ax,t^by,t^cz)$. (For example, p.3 in https://www.math.tecnico.ulisboa.pt/~mabreu/preprints/weighted.pdf) On the other hand, according to p.6, Example 7 of https://www.intlpress.com/site/pub/files/_fulltext/journals/pamq/2008/0004/0002/PAMQ-2008-0004-0002-a001.pdf, $P(a,b,c)$ is the quotient of $S^5$ by the $S^1$-action given by $t\cdot (x,y,z)=(t^{bc}x,t^{ca}y,t^{ab}z)$. If this is true then it means that $P(a,b,c)=P(bc,ca,ab)$, but I can't see why. How can we show that these two are homeomorphic?