If $a,b\in Z(G)$, $ord(a) = 24$, $ ord(b) = 21 $, $ ord(c) = 5$ prove that some element has order $210$

Question

Let$\DeclareMathOperator\ord{ord}\DeclareMathOperator\lcm{lcm}$ $G$ be a group and $Z= Z(G)$ denote its center. Suppose $a,b\in Z$ and $c\in G$ such that $\ord(a) = 24$, $\ord(b)=21$, $\ord(c)=5$. Prove that there exists an element in $G$ with order $210$.

What I tried

I know that $a,b,c$ are abelian with each other.

so if $\gcd(a,b,c) = n$ then

$(abc)^n = a^nb^nc^n$

I tried playing a lot with the $\lcm$ and $n$ but didn't get anywhere.

Any hint or solution will be appreciated.

Answer 1

Hint: $(a^{12}bc)^{210}=1$. Now what else do we need to make a conclusion here?